Section FS  Four Subsets
system:sage


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   <h3 class="likesectionHead"><a 
 id="x36-143000"></a>Section FS&#x00A0;&#x00A0;Four Subsets</h3>
<!--l. 393--><p class="noindent" ><a 
 id="section.FS"></a> From <a 
href="http://linear.ups.edu/" ><span 
class="cmti-12">A First Course in Linear Algebra</span></a><br 
class="newline" />Version 2.20<br 
class="newline" /><span 
class="cmsy-10x-x-120">&#x00A9;</span>&#x00A0;2004.<br 
class="newline" />Licensed under the <a 
href="http://www.gnu.org/licenses/fdl.html" >GNU Free Documentation License</a>.<br 
class="newline" /><a 
href="http://linear.ups.edu/" class="url" ><span 
class="cmtt-12">http://linear.ups.edu/</span></a><br 
class="newline" /><br 
class="newline" /><a 
 id="x36-143000doc"></a> <a 
 id="dx36-143001"></a> There are four natural subsets associated with a matrix. We have met
three already: the null space, the column space and the row space. In this
section we will introduce a fourth, the left null space. The objective of this
section is to describe one procedure that will allow us to find linearly
independent sets that span each of these four sets of column vectors.
Along the way, we will make a connection with the inverse of a matrix, so
<a 
href="#theorem.FS">Theorem&#x00A0;FS</a> will tie together most all of this chapter (and the entire course so
far).
</p><!--l. 19--><p class="noindent" >
</p>
   <h4 class="likesubsectionHead"><a 
 id="x36-144000"></a>Subsection LNS: Left Null Space</h4>
<!--l. 19--><p class="noindent" ><a 
 id="subsection.FS.LNS"></a> <a 
 id="x36-144000doc"></a><a 
 id="dx36-144001"></a>
</p><!--l. 22--><p class="noindent" ><span 
class="cmbx-12">Definition</span><span 
class="cmbx-12">&#x00A0;LNS</span><br 
class="newline" /><a 
 id="definition.LNS"><span 
class="cmbx-12">Left Null Space</span></a><a 
 id="dx36-144002"></a><a 
 id="dx36-144003"></a><a 
 id="dx36-144004"></a><br 
class="newline" /> Suppose <!--l. 23--><span class="math" 
>A</span> is an
<!--l. 23--><span class="math" 
>m &#x00D7; n</span> matrix. Then the <span 
class="cmbx-12">left</span>
                                                                          

                                                                          
<span 
class="cmbx-12">null space </span>is defined as <!--l. 23--><span class="math" 
>&#x2112;\kern -1.95872pt \left (A\right ) = N\kern -1.95872pt \left ({A}^{t}\right ) &#x2286; {&#x2102;}^{m}</span>.
<a 
 id="dx36-144005"></a><a 
 id="dx36-144006"></a><a 
 id="dx36-144007"></a>
</p><!--l. 24--><p class="noindent" >(This definition contains <a 
 id="notation.LNS">Notation LNS</a>.)
<!--l. 25--><span class="math" 
>&#x25B3;</span>
</p><!--l. 27--><p class="indent" >   The left null space will not feature prominently in the sequel, but
we can explain its name and connect it to row operations. Suppose
<!--l. 27--><span class="math" 
>y &#x2208;&#x2112;\kern -1.95872pt \left (A\right )</span>. Then by
<a 
href="#definition.LNS">Definition&#x00A0;LNS</a>, <!--l. 27--><span class="math" 
>{A}^{t}y = 0</span>.
We can then write
</p><!--tex4ht:inline--><!--l. 40--><div class="math" 
>\eqalignno{
                {0}^{t}                &amp; ={ \left ({A}^{t}y\right )}^{t}                 &amp;                &amp;\text{@(a 
href="#definition.LNS")Definition LNS@(/a)}                &amp;                &amp;                &amp;                &amp;
                \cr 
                  &amp; = {y}^{t}{\left ({A}^{t}\right )}^{t}                &amp;                &amp;\text{@(a 
href="fcla-jsmath-latestli31.html#theorem.MMT")Theorem MMT@(/a)}                &amp;                &amp;                &amp;                &amp;
                \cr 
                  &amp; = {y}^{t}A                   &amp;                &amp;\text{@(a 
href="fcla-jsmath-latestli30.html#theorem.TT")Theorem TT@(/a)}                  &amp;                &amp;                &amp;                &amp;
   }</div>
<!--l. 42--><p class="noindent" >The product <!--l. 42--><span class="math" 
>{y}^{t}A</span> can be viewed
as the components of <!--l. 42--><span class="math" 
>y</span>
acting as the scalars in a linear combination of the <span 
class="cmti-12">rows </span>of
<!--l. 42--><span class="math" 
>A</span>. And the result is
a &#x201C;row vector&#x201D;, <!--l. 42--><span class="math" 
>{0}^{t}</span>
that is totally zeros. When we apply a sequence of row operations to a matrix,
each row of the resulting matrix is some linear combination of the rows. These
observations tell us that the vectors in the left null space are scalars that record a
sequence of row operations that result in a row of zeros in the row-reduced version
of the matrix. We will see this idea more explicitly in the course of proving
<a 
href="#theorem.FS">Theorem&#x00A0;FS</a>.
                                                                          

                                                                          
</p><!--l. 44--><p class="noindent" ><span 
class="cmbx-12">Example</span><span 
class="cmbx-12">&#x00A0;LNS</span><br 
class="newline" /><a 
 id="example.LNS"><span 
class="cmbx-12">Left null space</span></a><a 
 id="dx36-144008"></a><a 
 id="dx36-144009"></a><a 
 id="dx36-144010"></a><br 
class="newline" /> We will find the left null space of </p><table class="equation-star"><tr><td>
<!--l. 47--><div class="math" 
>
A = \left [\array{ 
 1 &amp;&#x2212;3&amp;1\cr 
&#x2212;2 &amp; 1 &amp;1
\cr 
 1 &amp; 5 &amp;1\cr 
 9 &amp;&#x2212;4 &amp;0 }                                                                                         \right ]
</div></td></tr></table>
<!--l. 57--><p class="indent" >   We transpose <!--l. 57--><span class="math" 
>A</span>
and row-reduce, </p><table class="equation-star"><tr><td>
<!--l. 59--><div class="math" 
>{
A}^{t} = \left [\array{ 
 1 &amp;&#x2212;2&amp;1&amp; 9\cr 
&#x2212;3 &amp; 1 &amp;5 &amp;&#x2212;4
\cr 
 1 &amp; 1 &amp;1&amp; 0 }                                                                                      \right ]\mathop{\longrightarrow}\limits_{}^{\text{RREF}}\left [\array{ 
\text{1}&amp;0&amp;0&amp; 2\cr 
0&amp;\text{1 } &amp;0 &amp;&#x2212;3
\cr 
0&amp;0&amp;\text{1}&amp; 1 }                                                                                           \right ]
</div></td></tr></table>
<!--l. 74--><p class="indent" >   Applying <a 
href="#definition.LNS">Definition&#x00A0;LNS</a> and <a 
href="fcla-jsmath-latestli26.html#theorem.BNS">Theorem&#x00A0;BNS</a> we have </p><table class="equation-star"><tr><td>
                                                                          

                                                                          
<!--l. 76--><div class="math" 
>
&#x2112;\kern -1.95872pt \left (A\right ) = N\kern -1.95872pt \left ({A}^{t}\right ) = \left \langle \left \{\left [\array{ 
&#x2212;2\cr 
 3
\cr 
&#x2212;1\cr 
 1 }                                                                                              \right ]\right \}\right \rangle 
</div></td></tr></table>
<!--l. 83--><p class="indent" >   If you row-reduce <!--l. 83--><span class="math" 
>A</span>
you will discover one zero row in the reduced row-echelon form. This zero row is created
by a sequence of row operations, which in total amounts to a linear combination, with
scalars <!--l. 83--><span class="math" 
>{a}_{1} = &#x2212;2</span>,
<!--l. 83--><span class="math" 
>{a}_{2} = 3</span>,
<!--l. 83--><span class="math" 
>{a}_{3} = &#x2212;1</span> and
<!--l. 83--><span class="math" 
>{a}_{4} = 1</span>, on the rows
of <!--l. 83--><span class="math" 
>A</span> and which
results in the zero vector (check this!). So the components of the vector describing the left
null space of <!--l. 83--><span class="math" 
>A</span>
provide a relation of linear dependence on the rows of
<!--l. 83--><span class="math" 
>A</span>.
<!--l. 84--><span class="math" 
>&#x22A0;</span>
</p><!--l. 86--><p class="noindent" >
</p>
   <h4 class="likesubsectionHead"><a 
 id="x36-145000"></a>Subsection CRS: Computing Column Spaces</h4>
<!--l. 86--><p class="noindent" ><a 
 id="subsection.FS.CRS"></a> <a 
 id="x36-145000doc"></a><a 
 id="dx36-145001"></a>  We have three ways to build the column space of a matrix. First, we can use
just the definition, <a 
href="fcla-jsmath-latestli34.html#definition.CSM">Definition&#x00A0;CSM</a>, and express the column space as a span of
the columns of the matrix. A second approach gives us the column space as the
span of <span 
class="cmti-12">some </span>of the columns of the matrix, but this set is linearly independent
(<a 
href="fcla-jsmath-latestli34.html#theorem.BCS">Theorem&#x00A0;BCS</a>). Finally, we can transpose the matrix, row-reduce the transpose,
kick out zero rows, and transpose the remaining rows back into column vectors.
<a 
href="fcla-jsmath-latestli34.html#theorem.CSRST">Theorem&#x00A0;CSRST</a> and <a 
href="fcla-jsmath-latestli34.html#theorem.BRS">Theorem&#x00A0;BRS</a> tell us that the resulting vectors are
linearly independent and their span is the column space of the original
matrix.
</p><!--l. 90--><p class="indent" >   We will now demonstrate a fourth method by way of a rather complicated
example. Study this example carefully, but realize that its main purpose is to
motivate a theorem that simplifies much of the apparent complexity. So other
                                                                          

                                                                          
than an instructive exercise or two, the procedure we are about to describe will
not be a usual approach to computing a column space.
</p><!--l. 92--><p class="noindent" ><span 
class="cmbx-12">Example</span><span 
class="cmbx-12">&#x00A0;CSANS</span><br 
class="newline" /><a 
 id="example.CSANS"><span 
class="cmbx-12">Column space as null space</span></a><a 
 id="dx36-145002"></a><a 
 id="dx36-145003"></a><a 
 id="dx36-145004"></a><br 
class="newline" /> Lets find the column space of the matrix
<!--l. 93--><span class="math" 
>A</span> below
with a new approach. </p><table class="equation-star"><tr><td>
<!--l. 95--><div class="math" 
>
A = \left [\array{ 
 10 &amp; 0 &amp; 3 &amp;  8  &amp;  7\cr 
&#x2212;16 &amp;&#x2212;1 &amp;&#x2212;4 &amp;&#x2212;10 &amp;&#x2212;13
\cr 
 &#x2212;6 &amp; 1 &amp;&#x2212;3&amp; &#x2212;6 &amp; &#x2212;6\cr 
 0 &amp; 2 &amp;&#x2212;2 &amp; &#x2212;3 &amp; &#x2212;2
\cr 
 3  &amp; 0 &amp; 1 &amp;  2  &amp;  3\cr 
 &#x2212;1 &amp;&#x2212;1 &amp; 1 &amp; 1 &amp; 0 }                                                                            \right ]
</div></td></tr></table>
<!--l. 107--><p class="indent" >   By <a 
href="fcla-jsmath-latestli34.html#theorem.CSCS">Theorem&#x00A0;CSCS</a> we know that the column vector
<!--l. 107--><span class="math" 
>b</span> is in the column
space of <!--l. 107--><span class="math" 
>A</span> if and only
if the linear system <!--l. 107--><span class="math" 
>&#x2112;S\kern -1.95872pt \left (A,\kern 1.95872pt b\right )</span>
is consistent. So let&#x2019;s try to solve this system in full generality, using a vector
of variables for the vector of constants. In other words, which vectors
<!--l. 107--><span class="math" 
>b</span>
lead to consistent systems? Begin by forming the augmented matrix
<!--l. 107--><span class="math" 
>\left [\left .A\kern 1.95872pt \right \vert \kern 1.95872pt b\right ]</span> with a general
version of <!--l. 107--><span class="math" 
>b</span>,
</p><table class="equation-star"><tr><td>
                                                                          

                                                                          
<!--l. 109--><div class="math" 
>
\left [\left .A\kern 1.95872pt \right \vert \kern 1.95872pt b\right ] = \left [\array{ 
 10 &amp; 0 &amp; 3 &amp;  8  &amp;  7  &amp;{b}_{1}
\cr 
&#x2212;16&amp;&#x2212;1&amp;&#x2212;4&amp;&#x2212;10&amp;&#x2212;13&amp;{b}_{2}
\cr 
 &#x2212;6 &amp; 1 &amp;&#x2212;3&amp; &#x2212;6 &amp; &#x2212;6 &amp;{b}_{3}
\cr 
 0  &amp; 2 &amp;&#x2212;2&amp; &#x2212;3 &amp; &#x2212;2 &amp;{b}_{4}
\cr 
 3  &amp; 0 &amp; 1 &amp;  2  &amp;  3  &amp;{b}_{5}
\cr 
 &#x2212;1 &amp;&#x2212;1&amp; 1 &amp;  1  &amp;  0  &amp;{b}_{6} }                                                                                                   \right ]
</div></td></tr></table>
<!--l. 121--><p class="indent" >   To identify solutions we will row-reduce this matrix and bring it to reduced
row-echelon form. Despite the presence of variables in the last column, there is
nothing to stop us from doing this. Except our numerical routines on calculators
can&#x2019;t be used, and even some of the symbolic algebra routines do some
unexpected maneuvers with this computation. So do it by hand. Yes, it is a
bit of work. But worth it. We&#x2019;ll still be here when you get back. Notice
along the way that the row operations are <span 
class="cmti-12">exactly </span>the same ones you
would do if you were just row-reducing the coefficient matrix alone, say in
connection with a homogeneous system of equations. The column with the
<!--l. 121--><span class="math" 
>{b}_{i}</span> acts as a
sort of bookkeeping device. There are many different possibilities for the result,
depending on what order you choose to perform the row operations, but shortly we&#x2019;ll
all be on the same page. Here&#x2019;s one possibility (you can find this same result by doing
additional row operations with the fifth and sixth rows to remove any occurrences
of <!--l. 121--><span class="math" 
>{b}_{1}</span>
and <!--l. 121--><span class="math" 
>{b}_{2}</span>
from the first four rows of your result):
</p><!--tex4ht:inline--><!--l. 132--><div class="math" 
>\eqalignno{
              \left [\array{ 
\text{1}&amp;0&amp;0&amp;0&amp; 2 &amp;    {b}_{3} &#x2212; {b}_{4} + 2{b}_{5} &#x2212; {b}_{6}
\cr 
0&amp;\text{1}&amp;0&amp;0&amp;&#x2212;3&amp;&#x2212;2{b}_{3} + 3{b}_{4} &#x2212; 3{b}_{5} + 3{b}_{6}
\cr 
0&amp;0&amp;\text{1}&amp;0&amp; 1 &amp;   {b}_{3} + {b}_{4} + 3{b}_{5} + 3{b}_{6}
\cr 
0&amp;0&amp;0&amp;\text{1}&amp;&#x2212;2&amp;     &#x2212;2{b}_{3} + {b}_{4} &#x2212; 4{b}_{5}
\cr 
0&amp;0&amp;0&amp;0&amp; 0 &amp;{b}_{1} + 3{b}_{3} &#x2212; {b}_{4} + 3{b}_{5} + {b}_{6}
\cr 
0&amp;0&amp;0&amp;0&amp; 0 &amp; {b}_{2} &#x2212; 2{b}_{3} + {b}_{4} + {b}_{5} &#x2212; {b}_{6} }                                                                                 \right ]              &amp;              &amp;
   }</div>
                                                                          

                                                                          
<!--l. 134--><p class="noindent" >Our goal is to identify those vectors <!--l. 134--><span class="math" 
>b</span>
which make <!--l. 134--><span class="math" 
>&#x2112;S\kern -1.95872pt \left (A,\kern 1.95872pt b\right )</span>
consistent. By <a 
href="fcla-jsmath-latestli19.html#theorem.RCLS">Theorem&#x00A0;RCLS</a> we know that the consistent systems are precisely those
without a leading 1 in the last column. Are the expressions in the last column of rows
5 and 6 equal to zero, or are they leading 1&#x2019;s? The answer is: maybe. It depends
on <!--l. 134--><span class="math" 
>b</span>.
With a nonzero value for either of these expressions, we would scale
the row and produce a leading 1. So we get a consistent system, and
<!--l. 134--><span class="math" 
>b</span> is
in the column space, if and only if these two expressions are both
simultaneously zero. In other words, members of the column space of
<!--l. 134--><span class="math" 
>A</span> are exactly
those vectors <!--l. 134--><span class="math" 
>b</span>
that satisfy
</p><!--tex4ht:inline--><!--l. 139--><div class="math" 
>\eqalignno{
                        {b}_{1} + 3{b}_{3} &#x2212; {b}_{4} + 3{b}_{5} + {b}_{6}                        &amp; = 0                        &amp;                        &amp;
                        \cr 
{b}_{2} &#x2212; 2{b}_{3} + {b}_{4} + {b}_{5} &#x2212; {b}_{6}                         &amp; = 0                        &amp;                        &amp;
   }</div>
<!--l. 141--><p class="noindent" >Hmmm. Looks suspiciously like a homogeneous system of two equations with six
variables. If you&#x2019;ve been playing along (and we hope you have) then you may have
a slightly different system, but you should have just two equations. Form the
coefficient matrix and row-reduce (notice that the system above has a coefficient
matrix that is already in reduced row-echelon form). We should all be together
now with the same matrix, </p><table class="equation-star"><tr><td>
                                                                          

                                                                          
<!--l. 143--><div class="math" 
>
L = \left [\array{ 
\text{1}&amp;0&amp; 3 &amp;&#x2212;1&amp;3&amp; 1\cr 
0&amp;\text{1 } &amp;&#x2212;2 &amp; 1 &amp;1 &amp;&#x2212;1 }                                                                                  \right ]
</div></td></tr></table>
<!--l. 151--><p class="indent" >   So, <!--l. 151--><span class="math" 
>C\kern -1.95872pt \left (A\right ) = N\kern -1.95872pt \left (L\right )</span>
and we can apply <a 
href="fcla-jsmath-latestli26.html#theorem.BNS">Theorem&#x00A0;BNS</a> to obtain a linearly independent set to use in a
span construction, </p><table class="equation-star"><tr><td>
<!--l. 153--><div class="math" 
>
C\kern -1.95872pt \left (A\right ) = N\kern -1.95872pt \left (L\right ) = \left \langle \left \{\left [\array{ 
&#x2212;3\cr 
 2
\cr 
 1\cr 
 0
\cr 
 0\cr 
 0  }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
 1\cr 
&#x2212;1
\cr 
 0\cr 
 1
\cr 
 0\cr 
 0  }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
&#x2212;3\cr 
&#x2212;1
\cr 
 0\cr 
 0
\cr 
 1\cr 
 0  }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
&#x2212;1\cr 
 1
\cr 
 0\cr 
 0
\cr 
 0\cr 
 1  }                                                                                              \right ]\right \}\right \rangle 
</div></td></tr></table>
<!--l. 162--><p class="indent" >   Whew! As a postscript to this central example, you may wish to
convince yourself that the four vectors above really are elements
of the column space? Do they create consistent systems with
<!--l. 162--><span class="math" 
>A</span> as
coefficient matrix? Can you recognize the constant vector in your description of
these solution sets?
</p><!--l. 164--><p class="indent" >   OK, that was so much fun, let&#x2019;s do it again. But simpler this time. And
we&#x2019;ll all get the same results all the way through. Doing row operations
by hand with variables can be a bit error prone, so let&#x2019;s see if we can
improve the process some. Rather than row-reduce a column vector
<!--l. 164--><span class="math" 
>b</span> full of variables, let&#x2019;s
write <!--l. 164--><span class="math" 
>b = {I}_{6}b</span> and we will
                                                                          

                                                                          
row-reduce the matrix <!--l. 164--><span class="math" 
>{I}_{6}</span>
and when we finish row-reducing, <span 
class="cmti-12">then </span>we will compute the matrix-vector
product. You should first convince yourself that we can operate like this (this
is the subject of a future homework exercise). Rather than augmenting
<!--l. 166--><span class="math" 
>A</span> with
<!--l. 166--><span class="math" 
>b</span>, we will instead
augment it with <!--l. 166--><span class="math" 
>{I}_{6}</span>
(does this feel familiar?), </p><table class="equation-star"><tr><td>
<!--l. 168--><div class="math" 
>
M = \left [\array{ 
 10 &amp; 0 &amp; 3 &amp;  8  &amp;  7  &amp;1&amp;0&amp;0&amp;0&amp;0&amp;0\cr 
&#x2212;16 &amp;&#x2212;1 &amp;&#x2212;4 &amp;&#x2212;10 &amp;&#x2212;13 &amp;0 &amp;1 &amp;0 &amp;0 &amp;0 &amp;0
\cr 
 &#x2212;6 &amp; 1 &amp;&#x2212;3&amp; &#x2212;6 &amp; &#x2212;6 &amp;0&amp;0&amp;1&amp;0&amp;0&amp;0\cr 
 0 &amp; 2 &amp;&#x2212;2 &amp; &#x2212;3 &amp; &#x2212;2 &amp;0 &amp;0 &amp;0 &amp;1 &amp;0 &amp;0
\cr 
 3  &amp; 0 &amp; 1 &amp;  2  &amp;  3  &amp;0&amp;0&amp;0&amp;0&amp;1&amp;0\cr 
 &#x2212;1 &amp;&#x2212;1 &amp; 1 &amp; 1 &amp; 0 &amp;0 &amp;0 &amp;0 &amp;0 &amp;0 &amp;1 }                                                                 \right ]
</div></td></tr></table>
<!--l. 180--><p class="indent" >   We want to row-reduce the left-hand side of this matrix, but we will apply the
same row operations to the right-hand side as well. And once we get the left-hand
side in reduced row-echelon form, we will continue on to put leading 1&#x2019;s in the
final two rows, as well as clearing out the columns containing those two additional
leading 1&#x2019;s. It is these additional row operations that will ensure that we all get to
the same place, since the reduced row-echelon form is unique (<a 
href="fcla-jsmath-latestli18.html#theorem.RREFU">Theorem&#x00A0;RREFU</a>),
</p><table class="equation-star"><tr><td>
                                                                          

                                                                          
<!--l. 182--><div class="math" 
>
N = \left [\array{ 
1&amp;0&amp;0&amp;0&amp; 2 &amp;0&amp;0&amp; 1 &amp;&#x2212;1&amp; 2 &amp;&#x2212;1\cr 
0&amp;1 &amp;0 &amp;0 &amp;&#x2212;3 &amp;0 &amp;0 &amp;&#x2212;2 &amp; 3 &amp;&#x2212;3 &amp; 3
\cr 
0&amp;0&amp;1&amp;0&amp; 1 &amp;0&amp;0&amp; 1 &amp; 1 &amp; 3 &amp; 3\cr 
0&amp;0 &amp;0 &amp;1 &amp;&#x2212;2 &amp;0 &amp;0 &amp;&#x2212;2 &amp; 1 &amp;&#x2212;4 &amp; 0
\cr 
0&amp;0&amp;0&amp;0&amp; 0 &amp;1&amp;0&amp; 3 &amp;&#x2212;1&amp; 3 &amp; 1\cr 
0&amp;0 &amp;0 &amp;0 &amp; 0 &amp;0 &amp;1 &amp;&#x2212;2 &amp; 1 &amp; 1 &amp;&#x2212;1
   }                                                                                              \right ]
</div></td></tr></table>
<!--l. 194--><p class="indent" >   We are after the final six columns of this matrix, which we will multiply by
<!--l. 194--><span class="math" 
>b</span>
</p><table class="equation-star"><tr><td>
<!--l. 196--><div class="math" 
>
J = \left [\array{ 
0&amp;0&amp; 1 &amp;&#x2212;1&amp; 2 &amp;&#x2212;1\cr 
0&amp;0 &amp;&#x2212;2 &amp; 3 &amp;&#x2212;3 &amp; 3
\cr 
0&amp;0&amp; 1 &amp; 1 &amp; 3 &amp; 3\cr 
0&amp;0 &amp;&#x2212;2 &amp; 1 &amp;&#x2212;4 &amp; 0
\cr 
1&amp;0&amp; 3 &amp;&#x2212;1&amp; 3 &amp; 1\cr 
0&amp;1 &amp;&#x2212;2 &amp; 1 &amp; 1 &amp;&#x2212;1
   }                                                                                              \right ]
</div></td></tr></table>
<!--l. 208--><p class="indent" >   so </p><table class="equation-star"><tr><td>
                                                                          

                                                                          
<!--l. 210--><div class="math" 
>
Jb = \left [\array{ 
0&amp;0&amp; 1 &amp;&#x2212;1&amp; 2 &amp;&#x2212;1\cr 
0&amp;0 &amp;&#x2212;2 &amp; 3 &amp;&#x2212;3 &amp; 3
\cr 
0&amp;0&amp; 1 &amp; 1 &amp; 3 &amp; 3\cr 
0&amp;0 &amp;&#x2212;2 &amp; 1 &amp;&#x2212;4 &amp; 0
\cr 
1&amp;0&amp; 3 &amp;&#x2212;1&amp; 3 &amp; 1\cr 
0&amp;1 &amp;&#x2212;2 &amp; 1 &amp; 1 &amp;&#x2212;1
   }                                                                                              \right ]\left [\array{ 
{b}_{1}
\cr 
{b}_{2}
\cr 
{b}_{3}
\cr 
{b}_{4}
\cr 
{b}_{5}
\cr 
{b}_{6}   }                                                                                                                                  \right ] = \left [\array{ 
   {b}_{3} &#x2212; {b}_{4} + 2{b}_{5} &#x2212; {b}_{6}
\cr 
 &#x2212;2{b}_{3} + 3{b}_{4} &#x2212; 3{b}_{5} + 3{b}_{6}
\cr 
   {b}_{3} + {b}_{4} + 3{b}_{5} + 3{b}_{6}
\cr 
     &#x2212;2{b}_{3} + {b}_{4} &#x2212; 4{b}_{5}
\cr 
{b}_{1} + 3{b}_{3} &#x2212; {b}_{4} + 3{b}_{5} + {b}_{6}
\cr 
 {b}_{2} &#x2212; 2{b}_{3} + {b}_{4} + {b}_{5} &#x2212; {b}_{6}\cr 
 }                                                                                               \right ]
</div></td></tr></table>
<!--l. 232--><p class="indent" >   So by applying the same row operations that row-reduce
<!--l. 232--><span class="math" 
>A</span>
to the identity matrix (which we could do with a calculator once
<!--l. 232--><span class="math" 
>{I}_{6}</span> is placed
alongside of <!--l. 232--><span class="math" 
>A</span>),
we can then arrive at the result of row-reducing a column of symbols where
the vector of constants usually resides. Since the row-reduced version of
<!--l. 232--><span class="math" 
>A</span> has
two zero rows, for a consistent system we require that
</p><!--tex4ht:inline--><!--l. 237--><div class="math" 
>\eqalignno{
                        {b}_{1} + 3{b}_{3} &#x2212; {b}_{4} + 3{b}_{5} + {b}_{6}                        &amp; = 0                        &amp;                        &amp;
                        \cr 
{b}_{2} &#x2212; 2{b}_{3} + {b}_{4} + {b}_{5} &#x2212; {b}_{6}                         &amp; = 0                        &amp;                        &amp;
   }</div>
<!--l. 239--><p class="noindent" >Now we are exactly back where we were on the first go-round. Notice that we obtain the
matrix <!--l. 239--><span class="math" 
>L</span>
as simply the last two rows and last six columns of
<!--l. 239--><span class="math" 
>N</span>.
<!--l. 241--><span class="math" 
>&#x22A0;</span>
                                                                          

                                                                          
</p><!--l. 243--><p class="indent" >   This example motivates the remainder of this section, so it is worth careful
study. You might attempt to mimic the second approach with the coefficient
matrices of <a 
href="fcla-jsmath-latestli81.html#archetype.I">Archetype&#x00A0;I</a> and <a 
href="fcla-jsmath-latestli82.html#archetype.J">Archetype&#x00A0;J</a>. We will see shortly that the matrix
<!--l. 243--><span class="math" 
>L</span> contains more
information about <!--l. 243--><span class="math" 
>A</span>
than just the column space.
</p>
   <h4 class="likesubsectionHead"><a 
 id="x36-146000"></a>Subsection EEF: Extended echelon form</h4>
<!--l. 245--><p class="noindent" ><a 
 id="subsection.FS.EEF"></a>  <a 
 id="x36-146000doc"></a><a 
 id="dx36-146001"></a>  The final matrix that we row-reduced in <a 
href="#example.CSANS">Example&#x00A0;CSANS</a> should look
familiar in most respects to the procedure we used to compute the inverse of a
nonsingular matrix, <a 
href="fcla-jsmath-latestli32.html#theorem.CINM">Theorem&#x00A0;CINM</a>. We will now generalize that procedure to
matrices that are not necessarily nonsingular, or even square. First a
definition.
</p><!--l. 249--><p class="noindent" ><span 
class="cmbx-12">Definition</span><span 
class="cmbx-12">&#x00A0;EEF</span><br 
class="newline" /><a 
 id="definition.EEF"><span 
class="cmbx-12">Extended Echelon Form</span></a><a 
 id="dx36-146002"></a><a 
 id="dx36-146003"></a><a 
 id="dx36-146004"></a><br 
class="newline" /> Suppose <!--l. 250--><span class="math" 
>A</span> is an
<!--l. 250--><span class="math" 
>m &#x00D7; n</span> matrix. Add
<!--l. 250--><span class="math" 
>m</span> new columns to
<!--l. 250--><span class="math" 
>A</span> that together
equal an <!--l. 250--><span class="math" 
>m &#x00D7; m</span> identity
matrix to form an <!--l. 250--><span class="math" 
>m &#x00D7; (n + m)</span>
matrix <!--l. 250--><span class="math" 
>M</span>. Use row
operations to bring <!--l. 250--><span class="math" 
>M</span>
to reduced row-echelon form and call the result
<!--l. 250--><span class="math" 
>N</span>.
<!--l. 250--><span class="math" 
>N</span>
is the <span 
class="cmbx-12">extended reduced row-echelon form </span>of
<!--l. 250--><span class="math" 
>A</span>,
and we will standardize on names for five submatrices
(<!--l. 250--><span class="math" 
>B</span>,
<!--l. 250--><span class="math" 
>C</span>,
<!--l. 250--><span class="math" 
>J</span>,
<!--l. 250--><span class="math" 
>K</span>,
<!--l. 250--><span class="math" 
>L</span>) of
<!--l. 250--><span class="math" 
>N</span>.
                                                                          

                                                                          
</p><!--l. 252--><p class="indent" >   Let <!--l. 252--><span class="math" 
>B</span> denote the
<!--l. 252--><span class="math" 
>m &#x00D7; n</span> matrix formed
from the first <!--l. 252--><span class="math" 
>n</span>
columns of <!--l. 252--><span class="math" 
>N</span> and
let <!--l. 252--><span class="math" 
>J</span> denote the
<!--l. 252--><span class="math" 
>m &#x00D7; m</span> matrix formed
from the last <!--l. 252--><span class="math" 
>m</span>
columns of <!--l. 252--><span class="math" 
>N</span>.
Suppose that <!--l. 252--><span class="math" 
>B</span> has
<!--l. 252--><span class="math" 
>r</span> nonzero rows.
Further partition <!--l. 252--><span class="math" 
>N</span>
by letting <!--l. 252--><span class="math" 
>C</span>
denote the <!--l. 252--><span class="math" 
>r &#x00D7; n</span>
matrix formed from all of the non-zero rows of
<!--l. 252--><span class="math" 
>B</span>. Let
<!--l. 252--><span class="math" 
>K</span> be the
<!--l. 252--><span class="math" 
>r &#x00D7; m</span> matrix formed
from the first <!--l. 252--><span class="math" 
>r</span>
rows of <!--l. 252--><span class="math" 
>J</span>, while
<!--l. 252--><span class="math" 
>L</span> will be the
<!--l. 252--><span class="math" 
>(m &#x2212; r) &#x00D7; m</span> matrix formed
from the bottom <!--l. 252--><span class="math" 
>m &#x2212; r</span>
rows of <!--l. 252--><span class="math" 
>J</span>.
Pictorially, </p><table class="equation-star"><tr><td>
<!--l. 254--><div class="math" 
>
M = [A\vert {I}_{m}]\mathop{\longrightarrow}\limits_{}^{\text{RREF}}N = [B\vert J] = \left [\array{ 
C&amp;K<mtr 
class="hline"><mtd><mo> &#x0332; </mo></mtd> <mtd><mo> &#x0332; </mo></mtd> </mtr>\cr 
  0 &amp;L }                                                                                \right ]
</div></td></tr></table>
   <!--l. 262--><span class="math" 
>&#x25B3;</span>
<!--l. 265--><p class="noindent" ><span 
class="cmbx-12">Example</span><span 
class="cmbx-12">&#x00A0;SEEF</span><br 
class="newline" /><a 
 id="example.SEEF"><span 
class="cmbx-12">Submatrices of extended echelon form</span></a><a 
 id="dx36-146005"></a><a 
 id="dx36-146006"></a><a 
 id="dx36-146007"></a><br 
class="newline" /> We illustrate <a 
href="#definition.EEF">Definition&#x00A0;EEF</a> with the matrix
<!--l. 266--><span class="math" 
>A</span>,
</p><table class="equation-star"><tr><td>
<!--l. 268--><div class="math" 
>
A = \left [\array{ 
 1 &amp;&#x2212;1&amp;&#x2212;2&amp;  7  &amp; 1 &amp;  6\cr 
&#x2212;6 &amp; 2 &amp;&#x2212;4 &amp;&#x2212;18 &amp;&#x2212;3 &amp;&#x2212;26
\cr 
 4 &amp;&#x2212;1&amp; 4 &amp; 10 &amp; 2 &amp; 17\cr 
 3 &amp;&#x2212;1 &amp; 2 &amp; 9 &amp; 1 &amp; 12
   }                                                                                              \right ]
</div></td></tr></table>
<!--l. 278--><p class="indent" >   Augmenting with the <!--l. 278--><span class="math" 
>4 &#x00D7; 4</span>
identity matrix, M= </p><table class="equation-star"><tr><td>
<!--l. 281--><div class="math" 
>
\left [\array{ 
 1 &amp;&#x2212;1&amp;&#x2212;2&amp;  7  &amp; 1 &amp;  6  &amp;1&amp;0&amp;0&amp;0\cr 
&#x2212;6 &amp; 2 &amp;&#x2212;4 &amp;&#x2212;18 &amp;&#x2212;3 &amp;&#x2212;26 &amp;0 &amp;1 &amp;0 &amp;0
\cr 
 4 &amp;&#x2212;1&amp; 4 &amp; 10 &amp; 2 &amp; 17 &amp;0&amp;0&amp;1&amp;0\cr 
 3 &amp;&#x2212;1 &amp; 2 &amp; 9 &amp; 1 &amp; 12 &amp;0 &amp;0 &amp;0 &amp;1
   }                                                                                              \right ]
</div></td></tr></table>
<!--l. 290--><p class="indent" >   and row-reducing, we obtain </p><table class="equation-star"><tr><td>
                                                                          

                                                                          
<!--l. 292--><div class="math" 
>
N = \left [\array{ 
\text{1}&amp;0&amp;2&amp; 1 &amp;0&amp; 3 &amp;0&amp; 1 &amp;1&amp; 1\cr 
0&amp;\text{1 } &amp;4 &amp;&#x2212;6 &amp;0 &amp;&#x2212;1 &amp;0 &amp; 2 &amp;3 &amp; 0
\cr 
0&amp;0&amp;0&amp; 0 &amp;\text{1}&amp; 2 &amp;0&amp;&#x2212;1&amp;0&amp;&#x2212;2\cr 
0&amp;0 &amp;0 &amp; 0 &amp;0 &amp; 0 &amp;\text{1 } &amp; 2 &amp;2 &amp; 1 }                                                                          \right ]
</div></td></tr></table>
<!--l. 302--><p class="indent" >   So we then obtain
</p><!--tex4ht:inline--><!--l. 339--><div class="math" 
>\eqalignno{
                      B                       &amp; = \left [\array{ 
\text{1}&amp;0&amp;2&amp; 1 &amp;0&amp; 3\cr 
0&amp;\text{1 } &amp;4 &amp;&#x2212;6 &amp;0 &amp;&#x2212;1
\cr 
0&amp;0&amp;0&amp; 0 &amp;\text{1}&amp; 2\cr 
0&amp;0 &amp;0 &amp; 0 &amp;0 &amp; 0 }                                                                                      \right ]                      &amp;                      &amp;
                      \cr 
C                       &amp; = \left [\array{ 
\text{1}&amp;0&amp;2&amp; 1 &amp;0&amp; 3\cr 
0&amp;\text{1 } &amp;4 &amp;&#x2212;6 &amp;0 &amp;&#x2212;1
\cr 
0&amp;0&amp;0&amp; 0 &amp;\text{1}&amp; 2 }                                                                                      \right ]                      &amp;                      &amp;
                      \cr 
J                       &amp; = \left [\array{ 
0&amp; 1 &amp;1&amp; 1\cr 
0&amp; 2 &amp;3 &amp; 0
\cr 
0&amp;&#x2212;1&amp;0&amp;&#x2212;2\cr 
\text{1 }&amp; 2 &amp;2 &amp; 1 }                                                                                        \right ]                      &amp;                      &amp;
                      \cr 
K                      &amp; = \left [\array{ 
0&amp; 1 &amp;1&amp; 1\cr 
0&amp; 2 &amp;3 &amp; 0
\cr 
0&amp;&#x2212;1&amp;0&amp;&#x2212;2 }                                                                                        \right ]                      &amp;                      &amp;
                      \cr 
L                       &amp; = \left [\array{ 
\text{1}&amp;2&amp;2&amp;1 }                                                                                            \right ]                      &amp;                      &amp;
   }</div>
<!--l. 341--><p class="noindent" >You can observe (or verify) the properties of the following theorem with this example.
<!--l. 342--><span class="math" 
>&#x22A0;</span>
</p><!--l. 345--><p class="noindent" ><span 
class="cmbx-12">Theorem</span><span 
class="cmbx-12">&#x00A0;PEEF</span><br 
class="newline" /><a 
 id="theorem.PEEF"><span 
class="cmbx-12">Properties of Extended Echelon Form</span></a><a 
 id="dx36-146008"></a><a 
 id="dx36-146009"></a><a 
 id="dx36-146010"></a><br 
class="newline" /> Suppose that <!--l. 346--><span class="math" 
>A</span>
is an <!--l. 346--><span class="math" 
>m &#x00D7; n</span> matrix
and that <!--l. 346--><span class="math" 
>N</span>
                                                                          

                                                                          
is its extended echelon form. Then
     </p><ol  class="enumerate1" >
     <li 
  class="enumerate" id="x36-146012x1"><!--l. 349--><span class="math" 
>J</span>
     is nonsingular.
     </li>
     <li 
  class="enumerate" id="x36-146014x2"><!--l. 350--><span class="math" 
>B = JA</span>.
     </li>
     <li 
  class="enumerate" id="x36-146016x3">If <!--l. 351--><span class="math" 
>x &#x2208; {&#x2102;}^{n}</span>
     and <!--l. 351--><span class="math" 
>y &#x2208; {&#x2102;}^{m}</span>,
     then <!--l. 351--><span class="math" 
>Ax = y</span>
     if and only if <!--l. 351--><span class="math" 
>Bx = Jy</span>.
     </li>
     <li 
  class="enumerate" id="x36-146018x4"><!--l. 352--><span class="math" 
>C</span>
     is in reduced row-echelon form, has no zero rows and has <!--l. 352--><span class="math" 
>r</span>
     pivot columns.
     </li>
     <li 
  class="enumerate" id="x36-146020x5"><!--l. 353--><span class="math" 
>L</span>
     is in reduced row-echelon form, has no zero rows and has <!--l. 353--><span class="math" 
>m &#x2212; r</span>
     pivot columns.</li></ol>
<!--l. 356--><span class="math" 
>&#x25A1;</span>
<!--l. 358--><p class="noindent" ><span 
class="cmbx-12">Proof</span>&#x00A0;&#x00A0; <!--l. 359--><span class="math" 
>J</span>
is the result of applying a sequence of row operations to
<!--l. 359--><span class="math" 
>{I}_{m}</span>, as such
<!--l. 359--><span class="math" 
>J</span> and
<!--l. 359--><span class="math" 
>{I}_{m}</span> are row-equivalent.
<!--l. 359--><span class="math" 
>&#x2112;S\kern -1.95872pt \left ({I}_{m},\kern 1.95872pt 0\right )</span> has only the zero solution,
since <!--l. 359--><span class="math" 
>{I}_{m}</span> is nonsingular
(<a 
href="fcla-jsmath-latestli21.html#theorem.NMRRI">Theorem&#x00A0;NMRRI</a>). Thus, <!--l. 359--><span class="math" 
>&#x2112;S\kern -1.95872pt \left (J,\kern 1.95872pt 0\right )</span>
also has only the zero solution (<a 
href="fcla-jsmath-latestli18.html#theorem.REMES">Theorem&#x00A0;REMES</a>, <a 
href="fcla-jsmath-latestli17.html#definition.ESYS">Definition&#x00A0;ESYS</a>) and
<!--l. 359--><span class="math" 
>J</span> is
therefore nonsingular (<a 
href="fcla-jsmath-latestli20.html#definition.NSM">Definition&#x00A0;NSM</a>).
</p><!--l. 361--><p class="indent" >   To prove the second part of this conclusion, first convince yourself that row operations
and the matrix-vector are commutative operations. By this we mean the following. Suppose
                                                                          

                                                                          
that <!--l. 362--><span class="math" 
>F</span> is an
<!--l. 362--><span class="math" 
>m &#x00D7; n</span> matrix that is row-equivalent
to the matrix <!--l. 362--><span class="math" 
>G</span>. Apply
to the column vector <!--l. 362--><span class="math" 
>Fw</span>
the same sequence of row operations that converts
<!--l. 362--><span class="math" 
>F</span> to
<!--l. 362--><span class="math" 
>G</span>. Then the
result is <!--l. 362--><span class="math" 
>Gw</span>.
So we can do row operations on the matrix, then do a matrix-vector product, <span 
class="cmti-12">or</span>
do a matrix-vector product and then do row operations on a column vector, and
the result will be the same either way. Since matrix multiplication is defined by a
collection of matrix-vector products (<a 
href="fcla-jsmath-latestli31.html#definition.MM">Definition&#x00A0;MM</a>), if we apply to the matrix
product <!--l. 362--><span class="math" 
>FH</span>
the same sequence of row operations that converts
<!--l. 362--><span class="math" 
>F</span> to
<!--l. 362--><span class="math" 
>G</span> then the result will
equal <!--l. 362--><span class="math" 
>GH</span>. Now apply
these observations to <!--l. 362--><span class="math" 
>A</span>.
</p><!--l. 364--><p class="indent" >   Write <!--l. 364--><span class="math" 
>A{I}_{n} = {I}_{m}A</span> and apply the
row operations that convert <!--l. 364--><span class="math" 
>M</span>
to <!--l. 364--><span class="math" 
>N</span>.
<!--l. 364--><span class="math" 
>A</span> is converted
to <!--l. 364--><span class="math" 
>B</span>, while
<!--l. 364--><span class="math" 
>{I}_{m}</span> is converted
to <!--l. 364--><span class="math" 
>J</span>, so we
have <!--l. 364--><span class="math" 
>B{I}_{n} = JA</span>.
Simplifying the left side gives the desired conclusion.
</p><!--l. 366--><p class="indent" >   For the third conclusion, we now establish the two equivalences
</p><!--tex4ht:inline--><!--l. 374--><div class="math" 
>\eqalignno{
      Ax      &amp; = y      &amp;      &amp;\kern 3.26288pt \mathrel{&#x21D4;}\kern 3.26288pt       &amp;JAx      &amp; = Jy      &amp;      &amp;\kern 3.26288pt \mathrel{&#x21D4;}\kern 3.26288pt       &amp;Bx      &amp; = Jy      &amp;      &amp;      &amp;      &amp;      &amp;      &amp;      &amp;      &amp;      &amp;      &amp;
   }</div>
                                                                          

                                                                          
<!--l. 376--><p class="noindent" >The forward direction of the first equivalence is accomplished by multiplying both sides of the
matrix equality by <!--l. 376--><span class="math" 
>J</span>,
while the backward direction is accomplished by multiplying by the inverse of
<!--l. 376--><span class="math" 
>J</span> (which we know exists
by <a 
href="fcla-jsmath-latestli33.html#theorem.NI">Theorem&#x00A0;NI</a> since <!--l. 376--><span class="math" 
>J</span>
is nonsingular). The second equivalence is obtained simply by the substitutions given
by <!--l. 376--><span class="math" 
>JA = B</span>.
</p><!--l. 378--><p class="indent" >   The first <!--l. 378--><span class="math" 
>r</span>
rows of <!--l. 378--><span class="math" 
>N</span>
are in reduced row-echelon form, since any contiguous collection of
rows taken from a matrix in reduced row-echelon form will form a
matrix that is again in reduced row-echelon form. Since the matrix
<!--l. 378--><span class="math" 
>C</span> is formed by
removing the last <!--l. 378--><span class="math" 
>n</span>
entries of each these rows, the remainder is still in reduced row-echelon form. By its
construction, <!--l. 378--><span class="math" 
>C</span> has
no zero rows. <!--l. 378--><span class="math" 
>C</span>
has <!--l. 378--><span class="math" 
>r</span>
rows and each contains a leading 1, so there are
<!--l. 378--><span class="math" 
>r</span> pivot
columns in <!--l. 378--><span class="math" 
>C</span>.
</p><!--l. 380--><p class="indent" >   The final <!--l. 380--><span class="math" 
>m &#x2212; r</span>
rows of <!--l. 380--><span class="math" 
>N</span>
are in reduced row-echelon form, since any contiguous collection of
rows taken from a matrix in reduced row-echelon form will form a
matrix that is again in reduced row-echelon form. Since the matrix
<!--l. 380--><span class="math" 
>L</span> is formed by
removing the first <!--l. 380--><span class="math" 
>n</span>
entries of each these rows, and these entries are all zero (they form the zero rows
of <!--l. 380--><span class="math" 
>B</span>),
the remainder is still in reduced row-echelon form.
<!--l. 380--><span class="math" 
>L</span> is the final
<!--l. 380--><span class="math" 
>m &#x2212; r</span> rows of the
nonsingular matrix <!--l. 380--><span class="math" 
>J</span>,
so none of these rows can be totally zero, or
<!--l. 380--><span class="math" 
>J</span> would not row-reduce
                                                                          

                                                                          
to the identity matrix. <!--l. 380--><span class="math" 
>L</span>
has <!--l. 380--><span class="math" 
>m &#x2212; r</span>
rows and each contains a leading 1, so there are
<!--l. 380--><span class="math" 
>m &#x2212; r</span> pivot
columns in <!--l. 380--><span class="math" 
>L</span>.
   <!--l. 382--><span class="math" 
>&#x25A0;</span>
</p><!--l. 384--><p class="indent" >   Notice that in the case where <!--l. 384--><span class="math" 
>A</span>
is a nonsingular matrix we know that the reduced row-echelon form of
<!--l. 384--><span class="math" 
>A</span>
is the identity matrix (<a 
href="fcla-jsmath-latestli21.html#theorem.NMRRI">Theorem&#x00A0;NMRRI</a>), so
<!--l. 384--><span class="math" 
>B = {I}_{n}</span>. Then the second
conclusion above says <!--l. 384--><span class="math" 
>JA = B = {I}_{n}</span>,
so <!--l. 384--><span class="math" 
>J</span> is the
inverse of <!--l. 384--><span class="math" 
>A</span>.
Thus this theorem generalizes <a 
href="fcla-jsmath-latestli32.html#theorem.CINM">Theorem&#x00A0;CINM</a>, though the result is a &#x201C;left-inverse&#x201D;
of <!--l. 384--><span class="math" 
>A</span>
rather than a &#x201C;right-inverse.&#x201D;
</p><!--l. 386--><p class="indent" >   The third conclusion of <a 
href="#theorem.PEEF">Theorem&#x00A0;PEEF</a> is the most telling. It says that
<!--l. 386--><span class="math" 
>x</span> is a solution to the
linear system <!--l. 386--><span class="math" 
>&#x2112;S\kern -1.95872pt \left (A,\kern 1.95872pt y\right )</span> if and
only if <!--l. 386--><span class="math" 
>x</span> is a solution
to the linear system <!--l. 386--><span class="math" 
>&#x2112;S\kern -1.95872pt \left (B,\kern 1.95872pt Jy\right )</span>.
Or said differently, if we row-reduce the augmented matrix
<!--l. 386--><span class="math" 
>\left [\left .A\kern 1.95872pt \right \vert \kern 1.95872pt y\right ]</span> we will get the
augmented matrix <!--l. 386--><span class="math" 
>\left [\left .B\kern 1.95872pt \right \vert \kern 1.95872pt Jy\right ]</span>.
The matrix <!--l. 386--><span class="math" 
>J</span>
tracks the cumulative effect of the row operations that converts
<!--l. 386--><span class="math" 
>A</span>
to reduced row-echelon form, here effectively applying them
to the vector of constants in a system of equations having
<!--l. 386--><span class="math" 
>A</span> as a coefficient
matrix. When <!--l. 386--><span class="math" 
>A</span>
row-reduces to a matrix with zero rows, then
<!--l. 386--><span class="math" 
>Jy</span>
should also have zero entries in the same rows if the system is to be consistent.
                                                                          

                                                                          
</p><!--l. 388--><p class="noindent" >
</p>
   <h4 class="likesubsectionHead"><a 
 id="x36-147000"></a>Subsection FS: Four Subsets</h4>
<!--l. 388--><p class="noindent" ><a 
 id="subsection.FS.FS"></a> <a 
 id="x36-147000doc"></a><a 
 id="dx36-147001"></a>  With all the preliminaries in place we can state our main result for this
section. In essence this result will allow us to say that we can find linearly
independent sets to use in span constructions for all four subsets (null space,
column space, row space, left null space) by analyzing only the extended
echelon form of the matrix, and specifically, just the two submatrices
<!--l. 390--><span class="math" 
>C</span> and
<!--l. 390--><span class="math" 
>L</span>,
which will be ripe for analysis since they are already in reduced row-echelon form
(<a 
href="#theorem.PEEF">Theorem&#x00A0;PEEF</a>).
</p><!--l. 392--><p class="noindent" ><span 
class="cmbx-12">Theorem</span><span 
class="cmbx-12">&#x00A0;FS</span><br 
class="newline" /><a 
 id="theorem.FS"><span 
class="cmbx-12">Four Subsets</span></a><a 
 id="dx36-147002"></a><a 
 id="dx36-147003"></a><a 
 id="dx36-147004"></a><br 
class="newline" /> <a 
 id="dx36-147005"></a>Suppose <!--l. 394--><span class="math" 
>A</span> is an
<!--l. 394--><span class="math" 
>m &#x00D7; n</span> matrix with extended
echelon form <!--l. 394--><span class="math" 
>N</span>. Suppose the
reduced row-echelon form of <!--l. 394--><span class="math" 
>A</span>
has <!--l. 394--><span class="math" 
>r</span> nonzero rows.
Then <!--l. 394--><span class="math" 
>C</span> is the submatrix
of <!--l. 394--><span class="math" 
>N</span> formed from the
first <!--l. 394--><span class="math" 
>r</span> rows and the
first <!--l. 394--><span class="math" 
>n</span> columns and
<!--l. 394--><span class="math" 
>L</span> is the submatrix
of <!--l. 394--><span class="math" 
>N</span> formed from
the last <!--l. 394--><span class="math" 
>m</span> columns
and the last <!--l. 394--><span class="math" 
>m &#x2212; r</span>
rows. Then
     </p><ol  class="enumerate1" >
     <li 
  class="enumerate" id="x36-147007x1">The null space of <!--l. 397--><span class="math" 
>A</span>
     is the null space of <!--l. 397--><span class="math" 
>C</span>,
     <!--l. 397--><span class="math" 
>N\kern -1.95872pt \left (A\right ) = N\kern -1.95872pt \left (C\right )</span>.
     </li>
     <li 
  class="enumerate" id="x36-147009x2">The row space of <!--l. 398--><span class="math" 
>A</span>
     is the row space of <!--l. 398--><span class="math" 
>C</span>,
     <!--l. 398--><span class="math" 
>&#x211B;\kern -1.95872pt \left (A\right ) = &#x211B;\kern -1.95872pt \left (C\right )</span>.
                                                                          

                                                                          
     </li>
     <li 
  class="enumerate" id="x36-147011x3">The column space of <!--l. 399--><span class="math" 
>A</span>
     is the null space of <!--l. 399--><span class="math" 
>L</span>,
     <!--l. 399--><span class="math" 
>C\kern -1.95872pt \left (A\right ) = N\kern -1.95872pt \left (L\right )</span>.
     </li>
     <li 
  class="enumerate" id="x36-147013x4">The left null space of <!--l. 400--><span class="math" 
>A</span>
     is the row space of <!--l. 400--><span class="math" 
>L</span>,
     <!--l. 400--><span class="math" 
>&#x2112;\kern -1.95872pt \left (A\right ) = &#x211B;\kern -1.95872pt \left (L\right )</span>.</li></ol>
<!--l. 403--><span class="math" 
>&#x25A1;</span>
<!--l. 405--><p class="noindent" ><span 
class="cmbx-12">Proof</span>&#x00A0;&#x00A0; First, <!--l. 406--><span class="math" 
>N\kern -1.95872pt \left (A\right ) = N\kern -1.95872pt \left (B\right )</span> since
<!--l. 406--><span class="math" 
>B</span> is row-equivalent to
<!--l. 406--><span class="math" 
>A</span> (<a 
href="fcla-jsmath-latestli18.html#theorem.REMES">Theorem&#x00A0;REMES</a>).
The zero rows of <!--l. 406--><span class="math" 
>B</span>
represent equations that are always true in the homogeneous system
<!--l. 406--><span class="math" 
>&#x2112;S\kern -1.95872pt \left (B,\kern 1.95872pt 0\right )</span>, so
the removal of these equations will not change the solution set. Thus, in turn,
<!--l. 406--><span class="math" 
>N\kern -1.95872pt \left (B\right ) = N\kern -1.95872pt \left (C\right )</span>.
</p><!--l. 408--><p class="indent" >   Second, <!--l. 408--><span class="math" 
>&#x211B;\kern -1.95872pt \left (A\right ) = &#x211B;\kern -1.95872pt \left (B\right )</span> since
<!--l. 408--><span class="math" 
>B</span> is row-equivalent to
<!--l. 408--><span class="math" 
>A</span> (<a 
href="fcla-jsmath-latestli34.html#theorem.REMRS">Theorem&#x00A0;REMRS</a>).
The zero rows of <!--l. 408--><span class="math" 
>B</span>
contribute nothing to the span that is the row space of
<!--l. 408--><span class="math" 
>B</span>, so
the removal of these rows will not change the row space. Thus, in turn,
<!--l. 408--><span class="math" 
>&#x211B;\kern -1.95872pt \left (B\right ) = &#x211B;\kern -1.95872pt \left (C\right )</span>.
</p><!--l. 410--><p class="indent" >   Third, we prove the set equality <!--l. 410--><span class="math" 
>C\kern -1.95872pt \left (A\right ) = N\kern -1.95872pt \left (L\right )</span>
with <a 
href="fcla-jsmath-latestli70.html#definition.SE">Definition&#x00A0;SE</a>. Begin by showing that
<!--l. 410--><span class="math" 
>C\kern -1.95872pt \left (A\right ) &#x2286;N\kern -1.95872pt \left (L\right )</span>. Choose
<!--l. 410--><span class="math" 
>y &#x2208;C\kern -1.95872pt \left (A\right ) &#x2286; {&#x2102;}^{m}</span>. Then there exists
a vector <!--l. 410--><span class="math" 
>x &#x2208; {&#x2102;}^{n}</span> such that
<!--l. 410--><span class="math" 
>Ax = y</span> (<a 
href="fcla-jsmath-latestli34.html#theorem.CSCS">Theorem&#x00A0;CSCS</a>).
Then for <!--l. 410--><span class="math" 
>1 &#x2264; k &#x2264; m &#x2212; r</span>,
                                                                          

                                                                          
</p><!--tex4ht:inline--><!--l. 426--><div class="math" 
>\eqalignno{
          {\left [Ly\right ]}_{k}          &amp; ={ \left [Jy\right ]}_{r+k}          &amp;          &amp;\text{$L$ a submatrix of $J$}                    &amp;          &amp;          &amp;          &amp;
          \cr 
               &amp; ={ \left [Bx\right ]}_{r+k}          &amp;          &amp;\text{@(a 
href="#theorem.PEEF")Theorem PEEF@(/a)}                        &amp;          &amp;          &amp;          &amp;
          \cr 
               &amp; ={ \left [Ox\right ]}_{k}            &amp;          &amp;\text{Zero matrix a submatrix of $B$}          &amp;          &amp;          &amp;          &amp;
          \cr 
               &amp; ={ \left [0\right ]}_{k}              &amp;          &amp;\text{@(a 
href="fcla-jsmath-latestli31.html#theorem.MMZM")Theorem MMZM@(/a)}                      &amp;          &amp;          &amp;          &amp;
   }</div>
<!--l. 428--><p class="noindent" >So, for all <!--l. 428--><span class="math" 
>1 &#x2264; k &#x2264; m &#x2212; r</span>,
<!--l. 428--><span class="math" 
>{\left [Ly\right ]}_{k} ={ \left [0\right ]}_{k}</span>. So by <a 
href="fcla-jsmath-latestli23.html#definition.CVE">Definition&#x00A0;CVE</a>
we have <!--l. 428--><span class="math" 
>Ly = 0</span>
and thus <!--l. 428--><span class="math" 
>y &#x2208;N\kern -1.95872pt \left (L\right )</span>.
</p><!--l. 430--><p class="indent" >   Now, show that <!--l. 430--><span class="math" 
>N\kern -1.95872pt \left (L\right ) &#x2286;C\kern -1.95872pt \left (A\right )</span>.
Choose <!--l. 430--><span class="math" 
>y &#x2208;N\kern -1.95872pt \left (L\right ) &#x2286; {&#x2102;}^{m}</span>. Form
the vector <!--l. 430--><span class="math" 
>Ky &#x2208; {&#x2102;}^{r}</span>. The
linear system <!--l. 430--><span class="math" 
>&#x2112;S\kern -1.95872pt \left (C,\kern 1.95872pt Ky\right )</span> is
consistent since <!--l. 430--><span class="math" 
>C</span>
is in reduced row-echelon form and has no zero rows (<a 
href="#theorem.PEEF">Theorem&#x00A0;PEEF</a>). Let
<!--l. 430--><span class="math" 
>x &#x2208; {&#x2102;}^{n}</span> denote a
solution to <!--l. 430--><span class="math" 
>&#x2112;S\kern -1.95872pt \left (C,\kern 1.95872pt Ky\right )</span>.
</p><!--l. 432--><p class="indent" >   Then for <!--l. 432--><span class="math" 
>1 &#x2264; j &#x2264; r</span>,
</p><!--tex4ht:inline--><!--l. 445--><div class="math" 
>\eqalignno{
            {\left [Bx\right ]}_{j}            &amp; ={ \left [Cx\right ]}_{j}            &amp;            &amp;\text{$C$ a submatrix of $B$}                  &amp;            &amp;            &amp;            &amp;
            \cr 
                 &amp; ={ \left [Ky\right ]}_{j}            &amp;            &amp;\text{$x$ a solution to $&#x2112;S\kern -1.95872pt \left (C,\kern 1.95872pt Ky\right )$}            &amp;            &amp;            &amp;            &amp;
            \cr 
                 &amp; ={ \left [Jy\right ]}_{j}            &amp;            &amp;\text{$K$ a submatrix of $J$}                  &amp;            &amp;            &amp;            &amp;
            \cr 
                 &amp;                    &amp;            &amp;                                     &amp;
   }</div>
                                                                          

                                                                          
<!--l. 447--><p class="noindent" >And for <!--l. 447--><span class="math" 
>r + 1 &#x2264; k &#x2264; m</span>,
</p><!--tex4ht:inline--><!--l. 463--><div class="math" 
>\eqalignno{
          {\left [Bx\right ]}_{k}          &amp; ={ \left [Ox\right ]}_{k&#x2212;r}          &amp;          &amp;\text{Zero matrix a submatrix of $B$}          &amp;          &amp;          &amp;          &amp;
          \cr 
               &amp; ={ \left [0\right ]}_{k&#x2212;r}            &amp;          &amp;\text{@(a 
href="fcla-jsmath-latestli31.html#theorem.MMZM")Theorem MMZM@(/a)}                      &amp;          &amp;          &amp;          &amp;
          \cr 
               &amp; ={ \left [Ly\right ]}_{k&#x2212;r}          &amp;          &amp;\text{$y$ in $N\kern -1.95872pt \left (L\right )$}                             &amp;          &amp;          &amp;          &amp;
          \cr 
               &amp; ={ \left [Jy\right ]}_{k}            &amp;          &amp;\text{$L$ a submatrix of $J$}                    &amp;          &amp;          &amp;          &amp;
          \cr 
               &amp;                    &amp;          &amp;                                       &amp;
   }</div>
<!--l. 465--><p class="noindent" >So for all <!--l. 465--><span class="math" 
>1 &#x2264; i &#x2264; m</span>,
<!--l. 465--><span class="math" 
>{\left [Bx\right ]}_{i} ={ \left [Jy\right ]}_{i}</span> and by <a 
href="fcla-jsmath-latestli23.html#definition.CVE">Definition&#x00A0;CVE</a> we
have <!--l. 465--><span class="math" 
>Bx = Jy</span>. From <a 
href="#theorem.PEEF">Theorem&#x00A0;PEEF</a>
we know then that <!--l. 465--><span class="math" 
>Ax = y</span>,
and therefore <!--l. 465--><span class="math" 
>y &#x2208;C\kern -1.95872pt \left (A\right )</span>
(<a 
href="fcla-jsmath-latestli34.html#theorem.CSCS">Theorem&#x00A0;CSCS</a>). By <a 
href="fcla-jsmath-latestli70.html#definition.SE">Definition&#x00A0;SE</a> we now have
<!--l. 465--><span class="math" 
>C\kern -1.95872pt \left (A\right ) = N\kern -1.95872pt \left (L\right )</span>.
</p><!--l. 467--><p class="indent" >   Fourth, we prove the set equality <!--l. 467--><span class="math" 
>&#x2112;\kern -1.95872pt \left (A\right ) = &#x211B;\kern -1.95872pt \left (L\right )</span>
with <a 
href="fcla-jsmath-latestli70.html#definition.SE">Definition&#x00A0;SE</a>. Begin by showing that
<!--l. 467--><span class="math" 
>&#x211B;\kern -1.95872pt \left (L\right ) &#x2286;&#x2112;\kern -1.95872pt \left (A\right )</span>. Choose
<!--l. 467--><span class="math" 
>y &#x2208;&#x211B;\kern -1.95872pt \left (L\right ) &#x2286; {&#x2102;}^{m}</span>. Then there exists
a vector <!--l. 467--><span class="math" 
>w &#x2208; {&#x2102;}^{m&#x2212;r}</span> such that
<!--l. 467--><span class="math" 
>y = {L}^{t}w</span> (<a 
href="fcla-jsmath-latestli34.html#definition.RSM">Definition&#x00A0;RSM</a>,
<a 
href="fcla-jsmath-latestli34.html#theorem.CSCS">Theorem&#x00A0;CSCS</a>). Then for <!--l. 467--><span class="math" 
>1 &#x2264; i &#x2264; n</span>,
                                                                          

                                                                          
</p><!--tex4ht:inline--><!--l. 510--><div class="math" 
>\eqalignno{
       {\left [{A}^{t}y\right ]}_{
i}       &amp; ={ \mathop{&#x2211;
  }}_{k=1}^{m}{\left [{A}^{t}\right ]}_{
ik}{\left [y\right ]}_{k}                           &amp;       &amp;\text{@(a 
href="fcla-jsmath-latestli31.html#theorem.EMP")Theorem EMP@(/a)}           &amp;       &amp;       &amp;       &amp;
       \cr 
             &amp; ={ \mathop{&#x2211;
  }}_{k=1}^{m}{\left [{A}^{t}\right ]}_{
ik}{\left [{L}^{t}w\right ]}_{
k}                        &amp;       &amp;\text{Definition of $w$}           &amp;       &amp;       &amp;       &amp;
       \cr 
             &amp; ={ \mathop{&#x2211;
  }}_{k=1}^{m}{\left [{A}^{t}\right ]}_{
ik}{ \mathop{&#x2211;
  }}_{&#x2113;=1}^{m&#x2212;r}{\left [{L}^{t}\right ]}_{
k&#x2113;}{\left [w\right ]}_{&#x2113;}           &amp;       &amp;\text{@(a 
href="fcla-jsmath-latestli31.html#theorem.EMP")Theorem EMP@(/a)}           &amp;       &amp;       &amp;       &amp;
       \cr 
             &amp; ={ \mathop{&#x2211;
  }}_{k=1}^{m}{ \mathop{&#x2211;
  }}_{&#x2113;=1}^{m&#x2212;r}{\left [{A}^{t}\right ]}_{
ik}{\left [{L}^{t}\right ]}_{
k&#x2113;}{\left [w\right ]}_{&#x2113;}           &amp;       &amp;\text{@(a 
href="fcla-jsmath-latestli69.html#property.DCN")Property DCN@(/a)}            &amp;       &amp;       &amp;       &amp;
       \cr 
             &amp; ={ \mathop{&#x2211;
  }}_{&#x2113;=1}^{m&#x2212;r}{ \mathop{&#x2211;
  }}_{k=1}^{m}{\left [{A}^{t}\right ]}_{
ik}{\left [{L}^{t}\right ]}_{
k&#x2113;}{\left [w\right ]}_{&#x2113;}           &amp;       &amp;\text{@(a 
href="fcla-jsmath-latestli69.html#property.CACN")Property CACN@(/a)}          &amp;       &amp;       &amp;       &amp;
       \cr 
             &amp; ={ \mathop{&#x2211;
  }}_{&#x2113;=1}^{m&#x2212;r}\left ({\mathop{&#x2211;
  }}_{k=1}^{m}{\left [{A}^{t}\right ]}_{
ik}{\left [{L}^{t}\right ]}_{
k&#x2113;}\right ){\left [w\right ]}_{&#x2113;}          &amp;       &amp;\text{@(a 
href="fcla-jsmath-latestli69.html#property.DCN")Property DCN@(/a)}            &amp;       &amp;       &amp;       &amp;
       \cr 
             &amp; ={ \mathop{&#x2211;
  }}_{&#x2113;=1}^{m&#x2212;r}\left ({\mathop{&#x2211;
  }}_{k=1}^{m}{\left [{A}^{t}\right ]}_{
ik}{\left [{J}^{t}\right ]}_{
k,r+&#x2113;}\right ){\left [w\right ]}_{&#x2113;}       &amp;       &amp;\text{$L$ a submatrix of $J$}       &amp;       &amp;       &amp;       &amp;
       \cr 
             &amp; ={ \mathop{&#x2211;
  }}_{&#x2113;=1}^{m&#x2212;r}{\left [{A}^{t}{J}^{t}\right ]}_{
i,r+&#x2113;}{\left [w\right ]}_{&#x2113;}                    &amp;       &amp;\text{@(a 
href="fcla-jsmath-latestli31.html#theorem.EMP")Theorem EMP@(/a)}           &amp;       &amp;       &amp;       &amp;
       \cr 
             &amp; ={ \mathop{&#x2211;
  }}_{&#x2113;=1}^{m&#x2212;r}{\left [{\left (JA\right )}^{t}\right ]}_{
i,r+&#x2113;}{\left [w\right ]}_{&#x2113;}                   &amp;       &amp;\text{@(a 
href="fcla-jsmath-latestli31.html#theorem.MMT")Theorem MMT@(/a)}           &amp;       &amp;       &amp;       &amp;
       \cr 
             &amp; ={ \mathop{&#x2211;
  }}_{&#x2113;=1}^{m&#x2212;r}{\left [{B}^{t}\right ]}_{
i,r+&#x2113;}{\left [w\right ]}_{&#x2113;}                      &amp;       &amp;\text{@(a 
href="#theorem.PEEF")Theorem PEEF@(/a)}          &amp;       &amp;       &amp;       &amp;
       \cr 
             &amp; ={ \mathop{&#x2211;
  }}_{&#x2113;=1}^{m&#x2212;r}0{\left [w\right ]}_{
&#x2113;}                             &amp;       &amp;\text{Zero rows in $B$}           &amp;       &amp;       &amp;       &amp;
       \cr 
             &amp; = 0                                          &amp;       &amp;\text{@(a 
href="fcla-jsmath-latestli69.html#property.ZCN")Property ZCN@(/a)}            &amp;       &amp;       &amp;       &amp;
       \cr 
             &amp; ={ \left [0\right ]}_{i}                                         &amp;       &amp;\text{@(a 
href="fcla-jsmath-latestli18.html#definition.ZCV")Definition ZCV@(/a)}           &amp;       &amp;       &amp;       &amp;
   }</div>
<!--l. 512--><p class="noindent" >Since <!--l. 512--><span class="math" 
>{\left [{A}^{t}y\right ]}_{
i} ={ \left [0\right ]}_{i}</span> for
<!--l. 512--><span class="math" 
>1 &#x2264; i &#x2264; n</span>, <a 
href="fcla-jsmath-latestli23.html#definition.CVE">Definition&#x00A0;CVE</a>
implies that <!--l. 512--><span class="math" 
>{A}^{t}y = 0</span>.
This means that <!--l. 512--><span class="math" 
>y &#x2208;N\kern -1.95872pt \left ({A}^{t}\right )</span>.
</p><!--l. 514--><p class="indent" >   Now, show that <!--l. 514--><span class="math" 
>&#x2112;\kern -1.95872pt \left (A\right ) &#x2286;&#x211B;\kern -1.95872pt \left (L\right )</span>.
Choose <!--l. 514--><span class="math" 
>y &#x2208;&#x2112;\kern -1.95872pt \left (A\right ) &#x2286; {&#x2102;}^{m}</span>. The matrix
<!--l. 514--><span class="math" 
>J</span> is nonsingular
(<a 
href="#theorem.PEEF">Theorem&#x00A0;PEEF</a>), so <!--l. 514--><span class="math" 
>{J}^{t}</span>
                                                                          

                                                                          
is also nonsingular (<a 
href="fcla-jsmath-latestli32.html#theorem.MIT">Theorem&#x00A0;MIT</a>) and therefore the linear system
<!--l. 514--><span class="math" 
>&#x2112;S\kern -1.95872pt \left ({J}^{t},\kern 1.95872pt y\right )</span>
has a unique solution. Denote this solution as
<!--l. 514--><span class="math" 
>x &#x2208; {&#x2102;}^{m}</span>. We will need to work
with two &#x201C;halves&#x201D; of <!--l. 514--><span class="math" 
>x</span>,
which we will denote as <!--l. 514--><span class="math" 
>z</span>
and <!--l. 514--><span class="math" 
>w</span>
with formal definitions given by
</p><!--tex4ht:inline--><!--l. 525--><div class="math" 
>\eqalignno{
     {\left [z\right ]}_{j}     &amp; ={ \left [x\right ]}_{i}     &amp;     &amp;1 &#x2264; j &#x2264; r,     &amp;     &amp;     &amp;{\left [w\right ]}_{k}     &amp; ={ \left [x\right ]}_{r+k}     &amp;     &amp;1 &#x2264; k &#x2264; m &#x2212; r     &amp;     &amp;     &amp;     &amp;     &amp;     &amp;     &amp;     &amp;     &amp;     &amp;
   }</div>
<!--l. 527--><p class="noindent" >Now, for <!--l. 527--><span class="math" 
>1 &#x2264; j &#x2264; r</span>,
                                                                          

                                                                          
</p><!--tex4ht:inline--><!--l. 581--><div class="math" 
>\eqalignno{
    {\left [{C}^{t}z\right ]}_{
j}    &amp; ={ \mathop{&#x2211;
  }}_{k=1}^{r}{\left [{C}^{t}\right ]}_{
jk}{\left [z\right ]}_{k}                              &amp;    &amp;\text{@(a 
href="fcla-jsmath-latestli31.html#theorem.EMP")Theorem EMP@(/a)}            &amp;    &amp;    &amp;    &amp;
    \cr 
         &amp; ={ \mathop{&#x2211;
  }}_{k=1}^{r}{\left [{C}^{t}\right ]}_{
jk}{\left [z\right ]}_{k} +{ \mathop{&#x2211;
  }}_{&#x2113;=1}^{m&#x2212;r}{\left [O\right ]}_{
j&#x2113;}{\left [w\right ]}_{&#x2113;}        &amp;    &amp;\text{@(a 
href="fcla-jsmath-latestli30.html#definition.ZM")Definition ZM@(/a)}            &amp;    &amp;    &amp;    &amp;
    \cr 
         &amp; ={ \mathop{&#x2211;
  }}_{k=1}^{r}{\left [{B}^{t}\right ]}_{
jk}{\left [z\right ]}_{k} +{ \mathop{&#x2211;
  }}_{&#x2113;=1}^{m&#x2212;r}{\left [{B}^{t}\right ]}_{
j,r+&#x2113;}{\left [w\right ]}_{&#x2113;}     &amp;    &amp;\text{$C$, $O$ submatrices of $B$}    &amp;    &amp;    &amp;    &amp;
    \cr 
         &amp; ={ \mathop{&#x2211;
  }}_{k=1}^{r}{\left [{B}^{t}\right ]}_{
jk}{\left [x\right ]}_{k} +{ \mathop{&#x2211;
  }}_{&#x2113;=1}^{m&#x2212;r}{\left [{B}^{t}\right ]}_{
j,r+&#x2113;}{\left [x\right ]}_{r+&#x2113;}    &amp;    &amp;\text{Definitions of $z$ and $w$}    &amp;    &amp;    &amp;    &amp;
    \cr 
         &amp; ={ \mathop{&#x2211;
  }}_{k=1}^{r}{\left [{B}^{t}\right ]}_{
jk}{\left [x\right ]}_{k} +{ \mathop{&#x2211;
  }}_{k=r+1}^{m}{\left [{B}^{t}\right ]}_{
jk}{\left [x\right ]}_{k}       &amp;    &amp;\text{Re-index second sum}     &amp;    &amp;    &amp;    &amp;
    \cr 
         &amp; ={ \mathop{&#x2211;
  }}_{k=1}^{m}{\left [{B}^{t}\right ]}_{
jk}{\left [x\right ]}_{k}                             &amp;    &amp;\text{Combine sums}            &amp;    &amp;    &amp;    &amp;
    \cr 
         &amp; ={ \mathop{&#x2211;
  }}_{k=1}^{m}{\left [{\left (JA\right )}^{t}\right ]}_{
jk}{\left [x\right ]}_{k}                          &amp;    &amp;\text{@(a 
href="#theorem.PEEF")Theorem PEEF@(/a)}           &amp;    &amp;    &amp;    &amp;
    \cr 
         &amp; ={ \mathop{&#x2211;
  }}_{k=1}^{m}{\left [{A}^{t}{J}^{t}\right ]}_{
jk}{\left [x\right ]}_{k}                           &amp;    &amp;\text{@(a 
href="fcla-jsmath-latestli31.html#theorem.MMT")Theorem MMT@(/a)}           &amp;    &amp;    &amp;    &amp;
    \cr 
         &amp; ={ \mathop{&#x2211;
  }}_{k=1}^{m}{ \mathop{&#x2211;
  }}_{&#x2113;=1}^{m}{\left [{A}^{t}\right ]}_{
j&#x2113;}{\left [{J}^{t}\right ]}_{
&#x2113;k}{\left [x\right ]}_{k}                &amp;    &amp;\text{@(a 
href="fcla-jsmath-latestli31.html#theorem.EMP")Theorem EMP@(/a)}            &amp;    &amp;    &amp;    &amp;
    \cr 
         &amp; ={ \mathop{&#x2211;
  }}_{&#x2113;=1}^{m}{ \mathop{&#x2211;
  }}_{k=1}^{m}{\left [{A}^{t}\right ]}_{
j&#x2113;}{\left [{J}^{t}\right ]}_{
&#x2113;k}{\left [x\right ]}_{k}                &amp;    &amp;\text{@(a 
href="fcla-jsmath-latestli69.html#property.CACN")Property CACN@(/a)}          &amp;    &amp;    &amp;    &amp;
    \cr 
         &amp; ={ \mathop{&#x2211;
  }}_{&#x2113;=1}^{m}{\left [{A}^{t}\right ]}_{
j&#x2113;}\left ({\mathop{&#x2211;
  }}_{k=1}^{m}{\left [{J}^{t}\right ]}_{
&#x2113;k}{\left [x\right ]}_{k}\right )              &amp;    &amp;\text{@(a 
href="fcla-jsmath-latestli69.html#property.DCN")Property DCN@(/a)}            &amp;    &amp;    &amp;    &amp;
    \cr 
         &amp; ={ \mathop{&#x2211;
  }}_{&#x2113;=1}^{m}{\left [{A}^{t}\right ]}_{
j&#x2113;}{\left [{J}^{t}x\right ]}_{
&#x2113;}                           &amp;    &amp;\text{@(a 
href="fcla-jsmath-latestli31.html#theorem.EMP")Theorem EMP@(/a)}            &amp;    &amp;    &amp;    &amp;
    \cr 
         &amp; ={ \mathop{&#x2211;
  }}_{&#x2113;=1}^{m}{\left [{A}^{t}\right ]}_{
j&#x2113;}{\left [y\right ]}_{&#x2113;}                              &amp;    &amp;\text{Definition of $x$}            &amp;    &amp;    &amp;    &amp;
    \cr 
         &amp; ={ \left [{A}^{t}y\right ]}_{
j}                                        &amp;    &amp;\text{@(a 
href="fcla-jsmath-latestli31.html#theorem.EMP")Theorem EMP@(/a)}            &amp;    &amp;    &amp;    &amp;
    \cr 
         &amp; ={ \left [0\right ]}_{j}                                           &amp;    &amp;\text{$y &#x2208;&#x2112;\kern -1.95872pt \left (A\right )$}                 &amp;    &amp;    &amp;    &amp;
   }</div>
<!--l. 583--><p class="noindent" >So, by <a 
href="fcla-jsmath-latestli23.html#definition.CVE">Definition&#x00A0;CVE</a>, <!--l. 583--><span class="math" 
>{C}^{t}z = 0</span>
and the vector <!--l. 583--><span class="math" 
>z</span>
gives us a linear combination of the columns of
<!--l. 583--><span class="math" 
>{C}^{t}</span> that equals the zero
vector. In other words, <!--l. 583--><span class="math" 
>z</span>
gives a relation of linear dependence on the the rows of
<!--l. 583--><span class="math" 
>C</span>. However,
the rows of <!--l. 583--><span class="math" 
>C</span>
are a linearly independent set by <a 
href="fcla-jsmath-latestli34.html#theorem.BRS">Theorem&#x00A0;BRS</a>. According
                                                                          

                                                                          
to <a 
href="fcla-jsmath-latestli26.html#definition.LICV">Definition&#x00A0;LICV</a> we must conclude that the entries of
<!--l. 583--><span class="math" 
>z</span> are all
zero, i.e.&#x00A0;<!--l. 583--><span class="math" 
>z = 0</span>.
</p><!--l. 585--><p class="indent" >   Now, for <!--l. 585--><span class="math" 
>1 &#x2264; i &#x2264; m</span>,
we have
</p><!--tex4ht:inline--><!--l. 614--><div class="math" 
>\eqalignno{
     {\left [y\right ]}_{i}     &amp; ={ \left [{J}^{t}x\right ]}_{
i}                                        &amp;     &amp;\text{Definition of $x$}            &amp;     &amp;     &amp;     &amp;
     \cr 
         &amp; ={ \mathop{&#x2211;
  }}_{k=1}^{m}{\left [{J}^{t}\right ]}_{
ik}{\left [x\right ]}_{k}                             &amp;     &amp;\text{@(a 
href="fcla-jsmath-latestli31.html#theorem.EMP")Theorem EMP@(/a)}            &amp;     &amp;     &amp;     &amp;
     \cr 
         &amp; ={ \mathop{&#x2211;
  }}_{k=1}^{r}{\left [{J}^{t}\right ]}_{
ik}{\left [x\right ]}_{k} +{ \mathop{&#x2211;
  }}_{k=r+1}^{m}{\left [{J}^{t}\right ]}_{
ik}{\left [x\right ]}_{k}        &amp;     &amp;\text{Break apart sum}          &amp;     &amp;     &amp;     &amp;
     \cr 
         &amp; ={ \mathop{&#x2211;
  }}_{k=1}^{r}{\left [{J}^{t}\right ]}_{
ik}{\left [z\right ]}_{k} +{ \mathop{&#x2211;
  }}_{k=r+1}^{m}{\left [{J}^{t}\right ]}_{
ik}{\left [w\right ]}_{k&#x2212;r}     &amp;     &amp;\text{Definition of $z$ and $w$}     &amp;     &amp;     &amp;     &amp;
     \cr 
         &amp; ={ \mathop{&#x2211;
  }}_{k=1}^{r}{\left [{J}^{t}\right ]}_{
ik}0 +{ \mathop{&#x2211;
  }}_{&#x2113;=1}^{m&#x2212;r}{\left [{J}^{t}\right ]}_{
i,r+&#x2113;}{\left [w\right ]}_{&#x2113;}        &amp;     &amp;\text{$z = 0$, re-index}            &amp;     &amp;     &amp;     &amp;
     \cr 
         &amp; = 0 +{ \mathop{&#x2211;
  }}_{&#x2113;=1}^{m&#x2212;r}{\left [{L}^{t}\right ]}_{
i,&#x2113;}{\left [w\right ]}_{&#x2113;}                      &amp;     &amp;\text{$L$ a submatrix of $J$}        &amp;     &amp;     &amp;     &amp;
     \cr 
         &amp; ={ \left [{L}^{t}w\right ]}_{
i}                                       &amp;     &amp;\text{@(a 
href="fcla-jsmath-latestli31.html#theorem.EMP")Theorem EMP@(/a)}            &amp;     &amp;     &amp;     &amp;
     \cr 
         &amp;                                                &amp;     &amp;                          &amp;
   }</div>
<!--l. 616--><p class="noindent" >So by <a 
href="fcla-jsmath-latestli23.html#definition.CVE">Definition&#x00A0;CVE</a>, <!--l. 616--><span class="math" 
>y = {L}^{t}w</span>.
The existence of <!--l. 616--><span class="math" 
>w</span>
implies that <!--l. 616--><span class="math" 
>y &#x2208;&#x211B;\kern -1.95872pt \left (L\right )</span>, and
therefore <!--l. 616--><span class="math" 
>&#x2112;\kern -1.95872pt \left (A\right ) &#x2286;&#x211B;\kern -1.95872pt \left (L\right )</span>. So by
<a 
href="fcla-jsmath-latestli70.html#definition.SE">Definition&#x00A0;SE</a> we have <!--l. 616--><span class="math" 
>&#x2112;\kern -1.95872pt \left (A\right ) = &#x211B;\kern -1.95872pt \left (L\right )</span>.
<!--l. 618--><span class="math" 
>&#x25A0;</span>
</p><!--l. 620--><p class="indent" >   The first two conclusions of this theorem are nearly trivial. But they set up a pattern
of results for <!--l. 620--><span class="math" 
>C</span>
                                                                          

                                                                          
that is reflected in the latter two conclusions about
<!--l. 620--><span class="math" 
>L</span>. In
total, they tell us that we can compute all four subsets just by finding null spaces
and row spaces. This theorem does not tell us exactly how to compute
these subsets, but instead simply expresses them as null spaces and row
spaces of matrices in reduced row-echelon form without any zero rows
(<!--l. 620--><span class="math" 
>C</span> and
<!--l. 620--><span class="math" 
>L</span>). A
linearly independent set that spans the null space of a matrix in reduced
row-echelon form can be found easily with <a 
href="fcla-jsmath-latestli26.html#theorem.BNS">Theorem&#x00A0;BNS</a>. It is an even easier
matter to find a linearly independent set that spans the row space of a matrix in
reduced row-echelon form with <a 
href="fcla-jsmath-latestli34.html#theorem.BRS">Theorem&#x00A0;BRS</a>, especially when there are no zero
rows present. So an application of <a 
href="#theorem.FS">Theorem&#x00A0;FS</a> is typically followed by two
applications each of <a 
href="fcla-jsmath-latestli26.html#theorem.BNS">Theorem&#x00A0;BNS</a> and <a 
href="fcla-jsmath-latestli34.html#theorem.BRS">Theorem&#x00A0;BRS</a>.
</p><!--l. 622--><p class="indent" >   The situation when <!--l. 622--><span class="math" 
>r = m</span>
deserves comment, since now the matrix
<!--l. 622--><span class="math" 
>L</span> has no rows.
What is <!--l. 622--><span class="math" 
>C\kern -1.95872pt \left (A\right )</span>
when we try to apply <a 
href="#theorem.FS">Theorem&#x00A0;FS</a> and encounter
<!--l. 622--><span class="math" 
>N\kern -1.95872pt \left (L\right )</span>? One interpretation of
this situation is that <!--l. 622--><span class="math" 
>L</span>
is the coefficient matrix of a homogeneous system that has no equations. How
hard is it to find a solution vector to this system? Some thought will convince
you that <span 
class="cmti-12">any </span>proposed vector will qualify as a solution, since it makes
<span 
class="cmti-12">all </span>of the equations true. So every possible vector is in the null space of
<!--l. 622--><span class="math" 
>L</span> and
therefore <!--l. 622--><span class="math" 
>C\kern -1.95872pt \left (A\right ) = N\kern -1.95872pt \left (L\right ) = {&#x2102;}^{m}</span>.
OK, perhaps this sounds like some twisted argument from <span 
class="cmsl-12">Alice in Wonderland</span>.
Let us try another argument that might solidly convince you of this logic.
</p><!--l. 624--><p class="indent" >   If <!--l. 624--><span class="math" 
>r = m</span>,
when we row-reduce the augmented matrix of
<!--l. 624--><span class="math" 
>&#x2112;S\kern -1.95872pt \left (A,\kern 1.95872pt b\right )</span> the
result will have no zero rows, and all the leading 1&#x2019;s will occur in first
<!--l. 624--><span class="math" 
>n</span>
columns, so by <a 
href="fcla-jsmath-latestli19.html#theorem.RCLS">Theorem&#x00A0;RCLS</a> the system will be consistent. By <a 
href="fcla-jsmath-latestli34.html#theorem.CSCS">Theorem&#x00A0;CSCS</a>,
<!--l. 624--><span class="math" 
>b &#x2208;C\kern -1.95872pt \left (A\right )</span>.
Since <!--l. 624--><span class="math" 
>b</span>
was arbitrary, every possible vector is in the column space of
                                                                          

                                                                          
<!--l. 624--><span class="math" 
>A</span>, so we again have
<!--l. 624--><span class="math" 
>C\kern -1.95872pt \left (A\right ) = {&#x2102;}^{m}</span>. The situation
when a matrix has <!--l. 624--><span class="math" 
>r = m</span>
is known by the term <span 
class="cmbx-12">full rank</span>, and in the case of a square matrix coincides with
nonsingularity (see <a 
href="#exercise.FS.M50">Exercise&#x00A0;FS.M50</a>).
</p><!--l. 626--><p class="indent" >   The properties of the matrix <!--l. 626--><span class="math" 
>L</span>
described by this theorem can be explained informally as follows. A column vector
<!--l. 626--><span class="math" 
>y &#x2208; {&#x2102;}^{m}</span> is in the column
space of <!--l. 626--><span class="math" 
>A</span> if the
linear system <!--l. 626--><span class="math" 
>&#x2112;S\kern -1.95872pt \left (A,\kern 1.95872pt y\right )</span>
is consistent (<a 
href="fcla-jsmath-latestli34.html#theorem.CSCS">Theorem&#x00A0;CSCS</a>). By <a 
href="fcla-jsmath-latestli19.html#theorem.RCLS">Theorem&#x00A0;RCLS</a>, the reduced row-echelon form of the
augmented matrix <!--l. 626--><span class="math" 
>\left [\left .A\kern 1.95872pt \right \vert \kern 1.95872pt y\right ]</span>
of a consistent system will have zeros in the bottom
<!--l. 626--><span class="math" 
>m &#x2212; r</span>
locations of the last column. By <a 
href="#theorem.PEEF">Theorem&#x00A0;PEEF</a> this final column is the vector
<!--l. 626--><span class="math" 
>Jy</span> and so should then
have zeros in the final <!--l. 626--><span class="math" 
>m &#x2212; r</span>
locations. But since <!--l. 626--><span class="math" 
>L</span>
comprises the final <!--l. 626--><span class="math" 
>m &#x2212; r</span> rows
of <!--l. 626--><span class="math" 
>J</span>, this condition is
expressed by saying <!--l. 626--><span class="math" 
>y &#x2208;N\kern -1.95872pt \left (L\right )</span>.
</p><!--l. 628--><p class="indent" >   Additionally, the rows of <!--l. 628--><span class="math" 
>J</span>
are the scalars in linear combinations of the rows of
<!--l. 628--><span class="math" 
>A</span> that create the
rows of <!--l. 628--><span class="math" 
>B</span>. That
is, the rows of <!--l. 628--><span class="math" 
>J</span>
record the net effect of the sequence of row operations that takes
<!--l. 628--><span class="math" 
>A</span> to its reduced row-echelon
form, <!--l. 628--><span class="math" 
>B</span>. This can be seen in the
equation <!--l. 628--><span class="math" 
>JA = B</span> (<a 
href="#theorem.PEEF">Theorem&#x00A0;PEEF</a>).
As such, the rows of <!--l. 628--><span class="math" 
>L</span>
are scalars for linear combinations of the rows of
<!--l. 628--><span class="math" 
>A</span>
that yield zero rows. But such linear combinations are precisely the
elements of the left null space. So any element of the row space of
<!--l. 628--><span class="math" 
>L</span> is also an element of
the left null space of <!--l. 628--><span class="math" 
>A</span>.
We will now illustrate <a 
href="#theorem.FS">Theorem&#x00A0;FS</a> with a few examples.<br 
class="newline" />
</p><!--l. 633--><p class="noindent" ><span 
class="cmbx-12">Example</span><span 
class="cmbx-12">&#x00A0;FS1</span><br 
class="newline" /><a 
 id="example.FS1"><span 
class="cmbx-12">Four subsets, #1</span></a><a 
 id="dx36-147014"></a><a 
 id="dx36-147015"></a><a 
 id="dx36-147016"></a><br 
class="newline" /> In <a 
href="#example.SEEF">Example&#x00A0;SEEF</a> we found the five relevant submatrices of the matrix
</p><table class="equation-star"><tr><td>
<!--l. 636--><div class="math" 
>
A = \left [\array{ 
 1 &amp;&#x2212;1&amp;&#x2212;2&amp;  7  &amp; 1 &amp;  6\cr 
&#x2212;6 &amp; 2 &amp;&#x2212;4 &amp;&#x2212;18 &amp;&#x2212;3 &amp;&#x2212;26
\cr 
 4 &amp;&#x2212;1&amp; 4 &amp; 10 &amp; 2 &amp; 17\cr 
 3 &amp;&#x2212;1 &amp; 2 &amp; 9 &amp; 1 &amp; 12
   }                                                                                              \right ]
</div></td></tr></table>
<!--l. 646--><p class="indent" >   To apply <a 
href="#theorem.FS">Theorem&#x00A0;FS</a> we only need
<!--l. 646--><span class="math" 
>C</span> and
<!--l. 646--><span class="math" 
>L</span>,
</p><!--tex4ht:inline--><!--l. 660--><div class="math" 
>\eqalignno{
         C         &amp; = \left [\array{ 
\text{1}&amp;0&amp;2&amp; 1 &amp;0&amp; 3\cr 
0&amp;\text{1 } &amp;4 &amp;&#x2212;6 &amp;0 &amp;&#x2212;1
\cr 
0&amp;0&amp;0&amp; 0 &amp;\text{1}&amp; 2 }                                                                                      \right ]         &amp;L         &amp; = \left [\array{ 
\text{1}&amp;2&amp;2&amp;1 }                                                                                            \right ]         &amp;         &amp;         &amp;         &amp;
   }</div>
<!--l. 662--><p class="noindent" >Then we use <a 
href="#theorem.FS">Theorem&#x00A0;FS</a> to obtain
                                                                          

                                                                          
</p><!--tex4ht:inline--><!--l. 695--><div class="math" 
>\eqalignno{
         N\kern -1.95872pt \left (A\right )         &amp; = N\kern -1.95872pt \left (C\right ) = \left \langle \left \{\left [\array{ 
&#x2212;2\cr 
&#x2212;4
\cr 
 1\cr 
 0
\cr 
 0\cr 
 0  }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
&#x2212;1\cr 
 6
\cr 
 0\cr 
 1
\cr 
 0\cr 
 0  }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
&#x2212;3\cr 
 1
\cr 
 0\cr 
 0
\cr 
&#x2212;2\cr 
 1 }                                                                                              \right ]\right \}\right \rangle          &amp;         &amp;\text{@(a 
href="fcla-jsmath-latestli26.html#theorem.BNS")Theorem BNS@(/a)}         &amp;         &amp;         &amp;         &amp;
         \cr 
&#x211B;\kern -1.95872pt \left (A\right )         &amp; = &#x211B;\kern -1.95872pt \left (C\right ) = \left \langle \left \{\left [\array{ 
1\cr 
0
\cr 
2\cr 
1
\cr 
0\cr 
3   }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
 0\cr 
 1
\cr 
 4\cr 
&#x2212;6
\cr 
 0\cr 
&#x2212;1 }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
0\cr 
0
\cr 
0\cr 
0
\cr 
1\cr 
2   }                                                                                              \right ]\right \}\right \rangle          &amp;         &amp;\text{@(a 
href="fcla-jsmath-latestli34.html#theorem.BRS")Theorem BRS@(/a)}         &amp;         &amp;         &amp;         &amp;
         \cr 
C\kern -1.95872pt \left (A\right )          &amp; = N\kern -1.95872pt \left (L\right ) = \left \langle \left \{\left [\array{ 
&#x2212;2\cr 
 1
\cr 
 0\cr 
 0  }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
&#x2212;2\cr 
 0
\cr 
 1\cr 
 0  }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
&#x2212;1\cr 
 0
\cr 
 0\cr 
 1  }                                                                                              \right ]\right \}\right \rangle          &amp;         &amp;\text{@(a 
href="fcla-jsmath-latestli26.html#theorem.BNS")Theorem BNS@(/a)}         &amp;         &amp;         &amp;         &amp;
         \cr 
&#x2112;\kern -1.95872pt \left (A\right )          &amp; = &#x211B;\kern -1.95872pt \left (L\right ) = \left \langle \left \{\left [\array{ 
1\cr 
2
\cr 
2\cr 
1   }                                                                                              \right ]\right \}\right \rangle                                                                                                                                                                &amp;         &amp;\text{@(a 
href="fcla-jsmath-latestli34.html#theorem.BRS")Theorem BRS@(/a)}         &amp;         &amp;         &amp;         &amp;
   }</div>
<!--l. 697--><p class="noindent" >Boom! <!--l. 699--><span class="math" 
>&#x22A0;</span>
</p><!--l. 702--><p class="noindent" ><span 
class="cmbx-12">Example</span><span 
class="cmbx-12">&#x00A0;FS2</span><br 
class="newline" /><a 
 id="example.FS2"><span 
class="cmbx-12">Four subsets, #2</span></a><a 
 id="dx36-147017"></a><a 
 id="dx36-147018"></a><a 
 id="dx36-147019"></a><br 
class="newline" /> Now lets return to the matrix <!--l. 703--><span class="math" 
>A</span>
that we used to motivate this section in <a 
href="#example.CSANS">Example&#x00A0;CSANS</a>, </p><table class="equation-star"><tr><td>
                                                                          

                                                                          
<!--l. 705--><div class="math" 
>
A = \left [\array{ 
 10 &amp; 0 &amp; 3 &amp;  8  &amp;  7\cr 
&#x2212;16 &amp;&#x2212;1 &amp;&#x2212;4 &amp;&#x2212;10 &amp;&#x2212;13
\cr 
 &#x2212;6 &amp; 1 &amp;&#x2212;3&amp; &#x2212;6 &amp; &#x2212;6\cr 
 0 &amp; 2 &amp;&#x2212;2 &amp; &#x2212;3 &amp; &#x2212;2
\cr 
 3  &amp; 0 &amp; 1 &amp;  2  &amp;  3\cr 
 &#x2212;1 &amp;&#x2212;1 &amp; 1 &amp; 1 &amp; 0 }                                                                            \right ]
</div></td></tr></table>
<!--l. 717--><p class="indent" >   We form the matrix <!--l. 717--><span class="math" 
>M</span>
by adjoining the <!--l. 717--><span class="math" 
>6 &#x00D7; 6</span>
identity matrix <!--l. 717--><span class="math" 
>{I}_{6}</span>,
</p><table class="equation-star"><tr><td>
<!--l. 719--><div class="math" 
>
M = \left [\array{ 
 10 &amp; 0 &amp; 3 &amp;  8  &amp;  7  &amp;1&amp;0&amp;0&amp;0&amp;0&amp;0\cr 
&#x2212;16 &amp;&#x2212;1 &amp;&#x2212;4 &amp;&#x2212;10 &amp;&#x2212;13 &amp;0 &amp;1 &amp;0 &amp;0 &amp;0 &amp;0
\cr 
 &#x2212;6 &amp; 1 &amp;&#x2212;3&amp; &#x2212;6 &amp; &#x2212;6 &amp;0&amp;0&amp;1&amp;0&amp;0&amp;0\cr 
 0 &amp; 2 &amp;&#x2212;2 &amp; &#x2212;3 &amp; &#x2212;2 &amp;0 &amp;0 &amp;0 &amp;1 &amp;0 &amp;0
\cr 
 3  &amp; 0 &amp; 1 &amp;  2  &amp;  3  &amp;0&amp;0&amp;0&amp;0&amp;1&amp;0\cr 
 &#x2212;1 &amp;&#x2212;1 &amp; 1 &amp; 1 &amp; 0 &amp;0 &amp;0 &amp;0 &amp;0 &amp;0 &amp;1 }                                                                 \right ]
</div></td></tr></table>
<!--l. 731--><p class="indent" >   and row-reduce to obtain <!--l. 731--><span class="math" 
>N</span>
</p><table class="equation-star"><tr><td>
                                                                          

                                                                          
<!--l. 733--><div class="math" 
>
N = \left [\array{ 
\text{1}&amp;0&amp;0&amp;0&amp; 2 &amp;0&amp;0&amp; 1 &amp;&#x2212;1&amp; 2 &amp;&#x2212;1\cr 
0&amp;\text{1 } &amp;0 &amp;0 &amp;&#x2212;3 &amp;0 &amp;0 &amp;&#x2212;2 &amp; 3 &amp;&#x2212;3 &amp; 3
\cr 
0&amp;0&amp;\text{1}&amp;0&amp; 1 &amp;0&amp;0&amp; 1 &amp; 1 &amp; 3 &amp; 3\cr 
0&amp;0 &amp;0 &amp;\text{1 } &amp;&#x2212;2 &amp;0 &amp;0 &amp;&#x2212;2 &amp; 1 &amp;&#x2212;4 &amp; 0
\cr 
0&amp;0&amp;0&amp;0&amp; 0 &amp;\text{1}&amp;0&amp; 3 &amp;&#x2212;1&amp; 3 &amp; 1\cr 
0&amp;0 &amp;0 &amp;0 &amp; 0 &amp;0 &amp;\text{1 } &amp;&#x2212;2 &amp; 1 &amp; 1 &amp;&#x2212;1
   }                                                                                              \right ]
</div></td></tr></table>
<!--l. 745--><p class="indent" >   To find the four subsets for <!--l. 745--><span class="math" 
>A</span>,
we only need identify the <!--l. 745--><span class="math" 
>4 &#x00D7; 5</span>
matrix <!--l. 745--><span class="math" 
>C</span>
and the <!--l. 745--><span class="math" 
>2 &#x00D7; 6</span>
matrix <!--l. 745--><span class="math" 
>L</span>,
</p><!--tex4ht:inline--><!--l. 761--><div class="math" 
>\eqalignno{
      C      &amp; = \left [\array{ 
\text{1}&amp;0&amp;0&amp;0&amp; 2\cr 
0&amp;\text{1 } &amp;0 &amp;0 &amp;&#x2212;3
\cr 
0&amp;0&amp;\text{1}&amp;0&amp; 1\cr 
0&amp;0 &amp;0 &amp;\text{1 } &amp;&#x2212;2 }                                                                                        \right ]      &amp;L      &amp; = \left [\array{ 
\text{1}&amp;0&amp; 3 &amp;&#x2212;1&amp;3&amp; 1\cr 
0&amp;\text{1 } &amp;&#x2212;2 &amp; 1 &amp;1 &amp;&#x2212;1
   }                                                                                              \right ]      &amp;      &amp;      &amp;      &amp;
   }</div>
<!--l. 763--><p class="noindent" >Then we apply <a 
href="#theorem.FS">Theorem&#x00A0;FS</a>,
                                                                          

                                                                          
</p><!--tex4ht:inline--><!--l. 797--><div class="math" 
>\eqalignno{
       N\kern -1.95872pt \left (A\right )       &amp; = N\kern -1.95872pt \left (C\right ) = \left \langle \left \{\left [\array{ 
&#x2212;2\cr 
 3
\cr 
&#x2212;1\cr 
 2
\cr 
 1  }                                                                                              \right ]\right \}\right \rangle                                                                                                                                                                                                                                        &amp;       &amp;\text{@(a 
href="fcla-jsmath-latestli26.html#theorem.BNS")Theorem BNS@(/a)}       &amp;       &amp;       &amp;       &amp;
       \cr 
&#x211B;\kern -1.95872pt \left (A\right )        &amp; = &#x211B;\kern -1.95872pt \left (C\right ) = \left \langle \left \{\left [\array{ 
1\cr 
0
\cr 
0\cr 
0
\cr 
2   }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
 0\cr 
 1
\cr 
 0\cr 
 0
\cr 
&#x2212;3 }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
0\cr 
0
\cr 
1\cr 
0
\cr 
1   }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
 0\cr 
 0
\cr 
 0\cr 
 1
\cr 
&#x2212;2 }                                                                                              \right ]\right \}\right \rangle        &amp;       &amp;\text{@(a 
href="fcla-jsmath-latestli34.html#theorem.BRS")Theorem BRS@(/a)}       &amp;       &amp;       &amp;       &amp;
       \cr 
C\kern -1.95872pt \left (A\right )        &amp; = N\kern -1.95872pt \left (L\right ) = \left \langle \left \{\left [\array{ 
&#x2212;3\cr 
 2
\cr 
 1\cr 
 0
\cr 
 0\cr 
 0  }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
 1\cr 
&#x2212;1
\cr 
 0\cr 
 1
\cr 
 0\cr 
 0  }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
&#x2212;3\cr 
&#x2212;1
\cr 
 0\cr 
 0
\cr 
 1\cr 
 0  }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
&#x2212;1\cr 
 1
\cr 
 0\cr 
 0
\cr 
 0\cr 
 1  }                                                                                              \right ]\right \}\right \rangle        &amp;       &amp;\text{@(a 
href="fcla-jsmath-latestli26.html#theorem.BNS")Theorem BNS@(/a)}       &amp;       &amp;       &amp;       &amp;
       \cr 
&#x2112;\kern -1.95872pt \left (A\right )        &amp; = &#x211B;\kern -1.95872pt \left (L\right ) = \left \langle \left \{\left [\array{ 
 1\cr 
 0
\cr 
 3\cr 
&#x2212;1
\cr 
 3\cr 
 1  }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
 0\cr 
 1
\cr 
&#x2212;2\cr 
 1
\cr 
 1\cr 
&#x2212;1 }                                                                                              \right ]\right \}\right \rangle                                                                                                                                                              &amp;       &amp;\text{@(a 
href="fcla-jsmath-latestli34.html#theorem.BRS")Theorem BRS@(/a)}       &amp;       &amp;       &amp;       &amp;
   }</div>
   <!--l. 799--><span class="math" 
>&#x22A0;</span>
<!--l. 801--><p class="indent" >   The next example is just a bit different since the matrix has more rows than
columns, and a trivial null space.
</p><!--l. 803--><p class="noindent" ><span 
class="cmbx-12">Example</span><span 
class="cmbx-12">&#x00A0;FSAG</span><br 
class="newline" /><a 
 id="example.FSAG"><span 
class="cmbx-12">Four subsets, Archetype G</span></a><a 
 id="dx36-147020"></a><a 
 id="dx36-147021"></a><a 
 id="dx36-147022"></a><br 
class="newline" /> <a 
href="fcla-jsmath-latestli79.html#archetype.G">Archetype&#x00A0;G</a> and <a 
href="fcla-jsmath-latestli80.html#archetype.H">Archetype&#x00A0;H</a> are both systems of
<!--l. 805--><span class="math" 
>m = 5</span> equations
in <!--l. 805--><span class="math" 
>n = 2</span>
variables. They have identical coefficient matrices, which we will denote here as the
matrix <!--l. 805--><span class="math" 
>G</span>,
</p><table class="equation-star"><tr><td>
                                                                          

                                                                          
<!--l. 807--><div class="math" 
>
G = \left [\array{ 
 2 &amp; 3\cr 
&#x2212;1 &amp; 4
\cr 
 3 &amp;10\cr 
 3 &amp;&#x2212;1
\cr 
 6 &amp; 9 }                                                                                            \right ]
</div></td></tr></table>
<!--l. 812--><p class="indent" >   Adjoin the <!--l. 812--><span class="math" 
>5 &#x00D7; 5</span>
identity matrix, <!--l. 812--><span class="math" 
>{I}_{5}</span>,
to form </p><table class="equation-star"><tr><td>
<!--l. 814--><div class="math" 
>
M = \left [\array{ 
 2 &amp; 3 &amp;1&amp;0&amp;0&amp;0&amp;0\cr 
&#x2212;1 &amp; 4 &amp;0 &amp;1 &amp;0 &amp;0 &amp;0
\cr 
 3 &amp;10&amp;0&amp;0&amp;1&amp;0&amp;0\cr 
 3 &amp;&#x2212;1 &amp;0 &amp;0 &amp;0 &amp;1 &amp;0
\cr 
 6 &amp; 9 &amp;0&amp;0&amp;0&amp;0&amp;1 }                                                                                   \right ]
</div></td></tr></table>
<!--l. 825--><p class="indent" >   This row-reduces to </p><table class="equation-star"><tr><td>
<!--l. 827--><div class="math" 
>
N = \left [\array{ 
\text{1}&amp;0&amp;0&amp;0&amp;0&amp;  {3\over  _
11}  &amp;  {1\over  _
33}
\cr 
0&amp;\text{1}&amp;0&amp;0&amp;0&amp;&#x2212;{2\over _
11}&amp;  {1\over  _
11}
\cr 
0&amp;0&amp;\text{1}&amp;0&amp;0&amp;  0  &amp;&#x2212;{1\over 
3}
\cr 
0&amp;0&amp;0&amp;\text{1}&amp;0&amp;  1  &amp;&#x2212;{1\over 
3}
\cr 
0&amp;0&amp;0&amp;0&amp;\text{1}&amp;  1  &amp;&#x2212;1 }                                                                                  \right ]
</div></td></tr></table>
                                                                          

                                                                          
<!--l. 838--><p class="indent" >   The first <!--l. 838--><span class="math" 
>n = 2</span> columns
contain <!--l. 838--><span class="math" 
>r = 2</span> leading
1&#x2019;s, so we obtain <!--l. 838--><span class="math" 
>C</span>
as the <!--l. 838--><span class="math" 
>2 &#x00D7; 2</span> identity
matrix and extract <!--l. 838--><span class="math" 
>L</span>
from the final <!--l. 838--><span class="math" 
>m &#x2212; r = 3</span>
rows in the final <!--l. 838--><span class="math" 
>m = 5</span>
columns.
</p><!--tex4ht:inline--><!--l. 853--><div class="math" 
>\eqalignno{
            C            &amp; = \left [\array{ 
\text{1}&amp;0\cr 
0&amp;\text{1}  }                                                                                              \right ]            &amp;L            &amp; = \left [\array{ 
\text{1}&amp;0&amp;0&amp;0&amp;&#x2212;{1\over 
3}
\cr 
0&amp;\text{1}&amp;0&amp;1&amp;&#x2212;{1\over 
3}
\cr 
0&amp;0&amp;\text{1}&amp;1&amp;&#x2212;1 }                                                                                        \right ]            &amp;            &amp;            &amp;            &amp;
   }</div>
<!--l. 855--><p class="noindent" >Then we apply <a 
href="#theorem.FS">Theorem&#x00A0;FS</a>,
                                                                          

                                                                          
</p><!--tex4ht:inline--><!--l. 893--><div class="math" 
>\eqalignno{
         N\kern -1.95872pt \left (G\right ) = N\kern -1.95872pt \left (C\right )         &amp; = \left \langle &#x2205;\right \rangle  = \left \{0\right \}                                                                                                                                                                                                                                  &amp;         &amp;\text{@(a 
href="fcla-jsmath-latestli26.html#theorem.BNS")Theorem BNS@(/a)}         &amp;         &amp;         &amp;         &amp;
         \cr 
&#x211B;\kern -1.95872pt \left (G\right ) = &#x211B;\kern -1.95872pt \left (C\right )         &amp; = \left \langle \left \{\left [\array{ 
1\cr 
0   }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
0\cr 
1   }                                                                                              \right ]\right \}\right \rangle  = {&#x2102;}^{2}                                                                              &amp;         &amp;\text{@(a 
href="fcla-jsmath-latestli34.html#theorem.BRS")Theorem BRS@(/a)}         &amp;         &amp;         &amp;         &amp;
         \cr 
C\kern -1.95872pt \left (G\right ) = N\kern -1.95872pt \left (L\right )          &amp; = \left \langle \left \{\left [\array{ 
 0\cr 
&#x2212;1
\cr 
&#x2212;1\cr 
 1
\cr 
 0  }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
{1\over 
3}
\cr 
{1\over 
3}
\cr 
1\cr 
0
\cr 
1   }                                                                                              \right ]\right \}\right \rangle                                                                                    &amp;         &amp;\text{@(a 
href="fcla-jsmath-latestli26.html#theorem.BNS")Theorem BNS@(/a)}         &amp;         &amp;         &amp;         &amp;
         \cr 
                       &amp; = \left \langle \left \{\left [\array{ 
 0\cr 
&#x2212;1
\cr 
&#x2212;1\cr 
 1
\cr 
 0  }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
1\cr 
1
\cr 
3\cr 
0
\cr 
3   }                                                                                              \right ]\right \}\right \rangle                                                                                    &amp;         &amp;                       &amp;         &amp;
         \cr 
&#x2112;\kern -1.95872pt \left (G\right ) = &#x211B;\kern -1.95872pt \left (L\right )          &amp; = \left \langle \left \{\left [\array{ 
1\cr 
 0
\cr 
0\cr 
 0
\cr 
&#x2212;{1\over 
3} }                                                                                                                                  \right ],\kern 1.95872pt \left [\array{ 
0\cr 
 1
\cr 
0\cr 
 1
\cr 
&#x2212;{1\over 
3} }                                                                                                                                  \right ],\kern 1.95872pt \left [\array{ 
 0\cr 
 0
\cr 
 1\cr 
 1
\cr 
&#x2212;1 }                                                                                              \right ]\right \}\right \rangle          &amp;         &amp;\text{@(a 
href="fcla-jsmath-latestli34.html#theorem.BRS")Theorem BRS@(/a)}         &amp;         &amp;         &amp;         &amp;
         \cr 
                       &amp; = \left \langle \left \{\left [\array{ 
 3\cr 
 0
\cr 
 0\cr 
 0
\cr 
&#x2212;1 }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
 0\cr 
 3
\cr 
 0\cr 
 3\ 
\cr 
&#x2212;1 }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
 0\cr 
 0
\cr 
 1\cr 
 1
\cr 
&#x2212;1 }                                                                                              \right ]\right \}\right \rangle          &amp;         &amp;                       &amp;         &amp;
   }</div>
<!--l. 895--><p class="noindent" >As mentioned earlier, <a 
href="fcla-jsmath-latestli79.html#archetype.G">Archetype&#x00A0;G</a> is consistent, while <a 
href="fcla-jsmath-latestli80.html#archetype.H">Archetype&#x00A0;H</a> is
inconsistent. See if you can write the two different vectors of constants
from these two archetypes as linear combinations of the two vectors in
<!--l. 895--><span class="math" 
>C\kern -1.95872pt \left (G\right )</span>. How about the
two columns of <!--l. 895--><span class="math" 
>G</span>,
can you write each individually as a linear combination of the two vectors in
<!--l. 895--><span class="math" 
>C\kern -1.95872pt \left (G\right )</span>? They must be in
the column space of <!--l. 895--><span class="math" 
>G</span>
also. Are your answers unique? Do you notice anything about the
scalars that appear in the linear combinations you are forming?
<!--l. 897--><span class="math" 
>&#x22A0;</span>
</p><!--l. 899--><p class="indent" >   <a 
href="fcla-jsmath-latestli27.html#example.COV">Example&#x00A0;COV</a> and <a 
href="fcla-jsmath-latestli34.html#example.CSROI">Example&#x00A0;CSROI</a> each describes the column
space of the coefficient matrix from <a 
href="fcla-jsmath-latestli81.html#archetype.I">Archetype&#x00A0;I</a> as the span of a set of
<!--l. 899--><span class="math" 
>r = 3</span>
linearly independent vectors. It is no accident that these two different
sets both have the same size. If we (you?) were to calculate the
column space of this matrix using the null space of the matrix
                                                                          

                                                                          
<!--l. 899--><span class="math" 
>L</span> from
<a 
href="#theorem.FS">Theorem&#x00A0;FS</a> then we would again find a set of 3 linearly independent vectors that
span the range. More on this later.
</p><!--l. 901--><p class="indent" >   So we have three different methods to obtain a description of the column space
of a matrix as the span of a linearly independent set. <a 
href="fcla-jsmath-latestli34.html#theorem.BCS">Theorem&#x00A0;BCS</a> is sometimes
useful since the vectors it specifies are equal to actual columns of the matrix.
<a 
href="fcla-jsmath-latestli34.html#theorem.BRS">Theorem&#x00A0;BRS</a> and <a 
href="fcla-jsmath-latestli34.html#theorem.CSRST">Theorem&#x00A0;CSRST</a> combine to create vectors with lots of zeros,
and strategically placed 1&#x2019;s near the top of the vector. <a 
href="#theorem.FS">Theorem&#x00A0;FS</a> and the matrix
<!--l. 901--><span class="math" 
>L</span> from
the extended echelon form gives us a third method, which tends to create vectors
with lots of zeros, and strategically placed 1&#x2019;s near the bottom of the vector. If we
don&#x2019;t care about linear independence we can also appeal to <a 
href="fcla-jsmath-latestli34.html#definition.CSM">Definition&#x00A0;CSM</a> and
simply express the column space as the span of all the columns of the matrix,
giving us a fourth description.
</p><!--l. 903--><p class="indent" >   With <a 
href="fcla-jsmath-latestli34.html#theorem.CSRST">Theorem&#x00A0;CSRST</a> and <a 
href="fcla-jsmath-latestli34.html#definition.RSM">Definition&#x00A0;RSM</a>, we can compute column spaces
with theorems about row spaces, and we can compute row spaces with theorems
about row spaces, but in each case we must transpose the matrix first. At this
point you may be overwhelmed by all the possibilities for computing column and
row spaces. <a 
href="#diagram.CSRST">Diagram&#x00A0;CSRST</a> is meant to help. For both the column space and
row space, it suggests four techniques. One is to appeal to the definition, another
yields a span of a linearly independent set, and a third uses <a 
href="#theorem.FS">Theorem&#x00A0;FS</a>. A
fourth suggests transposing the matrix and the dashed line implies that then
the companion set of techniques can be applied. This can lead to a bit
of silliness, since if you were to follow the dashed lines <span 
class="cmti-12">twice </span>you would
transpose the matrix twice, and by <a 
href="fcla-jsmath-latestli30.html#theorem.TT">Theorem&#x00A0;TT</a> would accomplish nothing
productive.
<a 
 id="dx36-147023"></a>
<a 
 id="dx36-147024"></a>
</p>
<div class="center" 
>
<!--l. 905--><p class="noindent" >

</p><!--l. 905--><p class="noindent" ><img 
src="CSRST.png" alt="PIC"  
 /><br />
<a 
 id="diagram.CSRST">Diagram CSRST.   Column Space and Row Space Techniques</a>
</p>
                                                                          

                                                                          
</div>
<!--l. 907--><p class="noindent" >Although we have many ways to describe a column space, notice that one tempting
strategy will usually fail. It is not possible to simply row-reduce a matrix directly
and then use the columns of the row-reduced matrix as a set whose span equals
the column space. In other words, row operations <span 
class="cmti-12">do not </span>preserve column spaces
(however row operations do preserve row spaces, <a 
href="fcla-jsmath-latestli34.html#theorem.REMRS">Theorem&#x00A0;REMRS</a>). See
<a 
href="fcla-jsmath-latestli34.html#exercise.CRS.M21">Exercise&#x00A0;CRS.M21</a>.
</p><!--l. 393--><p class="noindent" >
</p>
   <h4 class="likesubsectionHead"><a 
 id="x36-148000"></a>Subsection READ: Reading Questions</h4>
<!--l. 393--><p class="noindent" ><a 
 id="subsection.FS.READ"></a> <a 
 id="x36-148000doc"></a><a 
 id="dx36-148001"></a>
     </p><ol  class="enumerate1" >
     <li 
  class="enumerate" id="x36-148003x1">Find a nontrivial element of the left null space of
     <!--l. 12--><span class="math" 
>A</span>.
     <table class="equation-star"><tr><td>
     <!--l. 14--><div class="math" 
>
A = \left [\array{ 
 2 &amp; 1 &amp;&#x2212;3&amp; 4\cr 
&#x2212;1 &amp;&#x2212;1 &amp; 2 &amp;&#x2212;1
\cr 
 0 &amp;&#x2212;1&amp; 1 &amp; 2}                                                                                    \right ]
</div></td></tr></table>
     </li>
     <li 
  class="enumerate" id="x36-148005x2">Find the matrices <!--l. 23--><span class="math" 
>C</span>
     and <!--l. 23--><span class="math" 
>L</span> in the extended
     echelon form of <!--l. 23--><span class="math" 
>A</span>.
     <table class="equation-star"><tr><td>
                                                                          

                                                                          
     <!--l. 25--><div class="math" 
>
A = \left [\array{ 
&#x2212;9&amp; 5 &amp;&#x2212;3\cr 
 2 &amp;&#x2212;1 &amp; 1
\cr 
&#x2212;5&amp; 3 &amp;&#x2212;1}                                                                                       \right ]
</div></td></tr></table>
     </li>
     <li 
  class="enumerate" id="x36-148007x3">Why is <a 
href="#theorem.FS">Theorem&#x00A0;FS</a> a great conclusion to <a 
href="fcla-jsmath-latestli29.html#chapter.M">Chapter&#x00A0;M</a>?</li></ol>
                                                                          

                                                                          
   <h4 class="likesubsectionHead"><a 
 id="x36-149000"></a>Subsection EXC: Exercises</h4>
<!--l. 393--><p class="noindent" ><a 
 id="subsection.FS.EXC"></a> <a 
 id="x36-149000doc"></a><a 
 id="dx36-149001"></a>  <a 
 id="exercise.FS.C20"><span 
class="cmbx-12">C20</span></a>   <a 
href="#example.FSAG">Example&#x00A0;FSAG</a> concludes with several questions. Perform the analysis
suggested by these questions. &#x00A0;<br 
class="newline" /> Contributed&#x00A0;by&#x00A0;<a 
href="fcla-jsmath-latestli6.html#BeezerRobert">Robert&#x00A0;Beezer</a>
</p><!--l. 11--><p class="noindent" ><a 
 id="exercise.FS.C25"><span 
class="cmbx-12">C25</span></a>   Given the matrix <!--l. 10--><span class="math" 
>A</span>
below, use the extended echelon form of
<!--l. 10--><span class="math" 
>A</span> to answer
each part of this problem. In each part, find a linearly independent set of vectors,
<!--l. 10--><span class="math" 
>S</span>, so that the
span of <!--l. 10--><span class="math" 
>S</span>,
<!--l. 10--><span class="math" 
>\left \langle S\right \rangle </span>,
equals the specified set of vectors. </p><table class="equation-star"><tr><td>
<!--l. 12--><div class="math" 
>
A = \left [\array{ 
&#x2212;5&amp; 3 &amp;&#x2212;1\cr 
&#x2212;1 &amp; 1 &amp; 1
\cr 
&#x2212;8&amp; 5 &amp;&#x2212;1\cr 
 3 &amp;&#x2212;2 &amp; 0
   }                                                                                              \right ]
</div></td></tr></table>
<!--l. 22--><p class="indent" >   (a)   The row space of <!--l. 22--><span class="math" 
>A</span>,
<!--l. 22--><span class="math" 
>&#x211B;\kern -1.95872pt \left (A\right )</span>.<br 
class="newline" />(b)   The column space of <!--l. 23--><span class="math" 
>A</span>,
<!--l. 23--><span class="math" 
>C\kern -1.95872pt \left (A\right )</span>.<br 
class="newline" />(c)   The null space of <!--l. 24--><span class="math" 
>A</span>,
<!--l. 24--><span class="math" 
>N\kern -1.95872pt \left (A\right )</span>.<br 
class="newline" />(d)   The left null space of <!--l. 25--><span class="math" 
>A</span>,
<!--l. 25--><span class="math" 
>&#x2112;\kern -1.95872pt \left (A\right )</span>.<br 
class="newline" />
</p><!--l. 11--><p class="indent" >   &#x00A0;<br 
class="newline" /> Contributed&#x00A0;by&#x00A0;<a 
href="fcla-jsmath-latestli6.html#BeezerRobert">Robert&#x00A0;Beezer</a>    <a 
href="#solution.FS.C25">Solution</a>&#x00A0;[<a 
href="#x36-150000doc">831<!--tex4ht:ref: solution.FS.C25 --></a>]
</p><!--l. 12--><p class="noindent" ><a 
 id="exercise.FS.C26"><span 
class="cmbx-12">C26</span></a>   For the matrix <!--l. 10--><span class="math" 
>D</span>
                                                                          

                                                                          
below use the extended echelon form to find<br 
class="newline" />(a)   a linearly independent set whose span is the column space of
<!--l. 11--><span class="math" 
>D</span>.<br 
class="newline" />(b)   a linearly independent set whose span is the left null space of
<!--l. 12--><span class="math" 
>D</span>.
</p><!--tex4ht:inline--><!--l. 22--><div class="math" 
>\eqalignno{
                        D                        &amp; = \left [\array{ 
&#x2212;7&amp;&#x2212;11&amp;&#x2212;19&amp;&#x2212;15\cr 
 6 &amp; 10 &amp; 18 &amp; 14
\cr 
 3 &amp;  5  &amp;  9  &amp;  7\cr 
&#x2212;1 &amp; &#x2212;2 &amp; &#x2212;4 &amp; &#x2212;3 }                                                                               \right ]                        &amp;                        &amp;
   }</div>
<!--l. 12--><p class="noindent" >&#x00A0;<br 
class="newline" /> Contributed&#x00A0;by&#x00A0;<a 
href="fcla-jsmath-latestli6.html#BeezerRobert">Robert&#x00A0;Beezer</a>    <a 
href="#solution.FS.C26">Solution</a>&#x00A0;[<a 
href="#x36-150000doc">833<!--tex4ht:ref: solution.FS.C26 --></a>]
</p><!--l. 13--><p class="noindent" ><a 
 id="exercise.FS.C41"><span 
class="cmbx-12">C41</span></a>   The following archetypes are systems of equations. For each system, write
the vector of constants as a linear combination of the vectors in the span
construction for the column space provided by <a 
href="#theorem.FS">Theorem&#x00A0;FS</a> and <a 
href="fcla-jsmath-latestli26.html#theorem.BNS">Theorem&#x00A0;BNS</a>
(these vectors are listed for each of these archetypes).<br 
class="newline" /><a 
href="fcla-jsmath-latestli73.html#archetype.A">Archetype&#x00A0;A</a><br 
class="newline" /><a 
href="fcla-jsmath-latestli74.html#archetype.B">Archetype&#x00A0;B</a><br 
class="newline" /><a 
href="fcla-jsmath-latestli75.html#archetype.C">Archetype&#x00A0;C</a><br 
class="newline" /><a 
href="fcla-jsmath-latestli76.html#archetype.D">Archetype&#x00A0;D</a><br 
class="newline" /><a 
href="fcla-jsmath-latestli77.html#archetype.E">Archetype&#x00A0;E</a><br 
class="newline" /><a 
href="fcla-jsmath-latestli78.html#archetype.F">Archetype&#x00A0;F</a><br 
class="newline" /><a 
href="fcla-jsmath-latestli79.html#archetype.G">Archetype&#x00A0;G</a><br 
class="newline" /><a 
href="fcla-jsmath-latestli80.html#archetype.H">Archetype&#x00A0;H</a><br 
class="newline" /><a 
href="fcla-jsmath-latestli81.html#archetype.I">Archetype&#x00A0;I</a><br 
class="newline" /><a 
href="fcla-jsmath-latestli82.html#archetype.J">Archetype&#x00A0;J</a>
</p><!--l. 13--><p class="indent" >   &#x00A0;<br 
class="newline" /> Contributed&#x00A0;by&#x00A0;<a 
href="fcla-jsmath-latestli6.html#BeezerRobert">Robert&#x00A0;Beezer</a>
                                                                          

                                                                          
</p><!--l. 14--><p class="noindent" ><a 
 id="exercise.FS.C43"><span 
class="cmbx-12">C43</span></a>   The following archetypes are either matrices or systems of equations with
coefficient matrices. For each matrix, compute the extended echelon form
<!--l. 10--><span class="math" 
>N</span> and identify
the matrices <!--l. 10--><span class="math" 
>C</span>
and <!--l. 10--><span class="math" 
>L</span>.
Using <a 
href="#theorem.FS">Theorem&#x00A0;FS</a>, <a 
href="fcla-jsmath-latestli26.html#theorem.BNS">Theorem&#x00A0;BNS</a> and <a 
href="fcla-jsmath-latestli34.html#theorem.BRS">Theorem&#x00A0;BRS</a> express the null space, the
row space, the column space and left null space of each coefficient matrix as a
span of a linearly independent set.<br 
class="newline" /><a 
href="fcla-jsmath-latestli73.html#archetype.A">Archetype&#x00A0;A</a><br 
class="newline" /><a 
href="fcla-jsmath-latestli74.html#archetype.B">Archetype&#x00A0;B</a><br 
class="newline" /><a 
href="fcla-jsmath-latestli75.html#archetype.C">Archetype&#x00A0;C</a><br 
class="newline" /><a 
href="fcla-jsmath-latestli76.html#archetype.D">Archetype&#x00A0;D</a>/<a 
href="fcla-jsmath-latestli77.html#archetype.E">Archetype&#x00A0;E</a><br 
class="newline" /><a 
href="fcla-jsmath-latestli78.html#archetype.F">Archetype&#x00A0;F</a><br 
class="newline" /><a 
href="fcla-jsmath-latestli79.html#archetype.G">Archetype&#x00A0;G</a>/<a 
href="fcla-jsmath-latestli80.html#archetype.H">Archetype&#x00A0;H</a><br 
class="newline" /><a 
href="fcla-jsmath-latestli81.html#archetype.I">Archetype&#x00A0;I</a><br 
class="newline" /><a 
href="fcla-jsmath-latestli82.html#archetype.J">Archetype&#x00A0;J</a><br 
class="newline" /><a 
href="fcla-jsmath-latestli83.html#archetype.K">Archetype&#x00A0;K</a><br 
class="newline" /><a 
href="fcla-jsmath-latestli84.html#archetype.L">Archetype&#x00A0;L</a>
</p><!--l. 14--><p class="indent" >   &#x00A0;<br 
class="newline" /> Contributed&#x00A0;by&#x00A0;<a 
href="fcla-jsmath-latestli6.html#BeezerRobert">Robert&#x00A0;Beezer</a>
</p><!--l. 15--><p class="noindent" ><a 
 id="exercise.FS.C60"><span 
class="cmbx-12">C60</span></a>   For the matrix <!--l. 10--><span class="math" 
>B</span>
below, find sets of vectors whose span equals the column space of
<!--l. 10--><span class="math" 
>B</span>
(<!--l. 10--><span class="math" 
>C\kern -1.95872pt \left (B\right )</span>) and
which individually meet the following extra requirements.<br 
class="newline" />(a)   The set illustrates the definition of the column space.<br 
class="newline" />(b)   The set is linearly independent and the members of the set are columns of
<!--l. 12--><span class="math" 
>B</span>.<br 
class="newline" />(c)   The set is linearly independent with a &#x201C;nice pattern of zeros and ones&#x201D; at the
<span 
class="cmti-12">top </span>of each vector.<br 
class="newline" />(d)   The set is linearly independent with a &#x201C;nice pattern of zeros and ones&#x201D; at
the bottom of each vector. </p><table class="equation-star"><tr><td>
                                                                          

                                                                          
<!--l. 16--><div class="math" 
>
B = \left [\array{ 
 2 &amp;3&amp;1&amp; 1\cr 
 1 &amp;1 &amp;0 &amp; 1
\cr 
&#x2212;1&amp;2&amp;3&amp;&#x2212;4 }                                                                                        \right ]
</div></td></tr></table>
<!--l. 15--><p class="indent" >   &#x00A0;<br 
class="newline" /> Contributed&#x00A0;by&#x00A0;<a 
href="fcla-jsmath-latestli6.html#BeezerRobert">Robert&#x00A0;Beezer</a>    <a 
href="#solution.FS.C60">Solution</a>&#x00A0;[<a 
href="#x36-150000doc">835<!--tex4ht:ref: solution.FS.C60 --></a>]
</p><!--l. 16--><p class="noindent" ><a 
 id="exercise.FS.C61"><span 
class="cmbx-12">C61</span></a>   Let <!--l. 10--><span class="math" 
>A</span>
be the matrix below, and find the indicated sets with the requested properties.
</p><table class="equation-star"><tr><td>
<!--l. 12--><div class="math" 
>
A = \left [\array{ 
 2 &amp;&#x2212;1&amp;  5  &amp;&#x2212;3\cr 
&#x2212;5 &amp; 3 &amp;&#x2212;12 &amp; 7
\cr 
 1 &amp; 1 &amp;  4  &amp;&#x2212;3 }                                                                                 \right ]
</div></td></tr></table>
<!--l. 21--><p class="indent" >   (a)   A linearly independent set <!--l. 21--><span class="math" 
>S</span>
so that <!--l. 21--><span class="math" 
>C\kern -1.95872pt \left (A\right ) = \left \langle S\right \rangle </span> and
<!--l. 21--><span class="math" 
>S</span> is composed of
columns of <!--l. 21--><span class="math" 
>A</span>.<br 
class="newline" />(b)   A linearly independent set <!--l. 22--><span class="math" 
>S</span>
so that <!--l. 22--><span class="math" 
>C\kern -1.95872pt \left (A\right ) = \left \langle S\right \rangle </span> and
the vectors in <!--l. 22--><span class="math" 
>S</span>
have a nice pattern of zeros and ones at the top of the vectors.<br 
class="newline" />(c)   A linearly independent set <!--l. 23--><span class="math" 
>S</span>
so that <!--l. 23--><span class="math" 
>C\kern -1.95872pt \left (A\right ) = \left \langle S\right \rangle </span> and
the vectors in <!--l. 23--><span class="math" 
>S</span>
have a nice pattern of zeros and ones at the bottom of the vectors.<br 
class="newline" />(d)   A linearly independent set <!--l. 24--><span class="math" 
>S</span>
so that <!--l. 24--><span class="math" 
>&#x211B;\kern -1.95872pt \left (A\right ) = \left \langle S\right \rangle </span>.
                                                                          

                                                                          
&#x00A0;<br 
class="newline" /> Contributed&#x00A0;by&#x00A0;<a 
href="fcla-jsmath-latestli6.html#BeezerRobert">Robert&#x00A0;Beezer</a>    <a 
href="#solution.FS.C61">Solution</a>&#x00A0;[<a 
href="#x36-150000doc">838<!--tex4ht:ref: solution.FS.C61 --></a>]
</p><!--l. 18--><p class="noindent" ><a 
 id="exercise.FS.M50"><span 
class="cmbx-12">M50</span></a>   Suppose that <!--l. 10--><span class="math" 
>A</span>
is a nonsingular matrix. Extend the four conclusions of <a 
href="#theorem.FS">Theorem&#x00A0;FS</a> in
this special case and discuss connections with previous results (such as
<a 
href="fcla-jsmath-latestli34.html#theorem.NME4">Theorem&#x00A0;NME4</a>). &#x00A0;<br 
class="newline" /> Contributed&#x00A0;by&#x00A0;<a 
href="fcla-jsmath-latestli6.html#BeezerRobert">Robert&#x00A0;Beezer</a>
</p><!--l. 19--><p class="noindent" ><a 
 id="exercise.FS.M51"><span 
class="cmbx-12">M51</span></a>   Suppose that <!--l. 10--><span class="math" 
>A</span>
is a singular matrix. Extend the four conclusions of <a 
href="#theorem.FS">Theorem&#x00A0;FS</a> in this special
case and discuss connections with previous results (such as <a 
href="fcla-jsmath-latestli34.html#theorem.NME4">Theorem&#x00A0;NME4</a>).
&#x00A0;<br 
class="newline" /> Contributed&#x00A0;by&#x00A0;<a 
href="fcla-jsmath-latestli6.html#BeezerRobert">Robert&#x00A0;Beezer</a>
                                                                          

                                                                          
</p>
   <h4 class="likesubsectionHead"><a 
 id="x36-150000"></a>Subsection SOL: Solutions</h4>
<!--l. 393--><p class="noindent" ><a 
 id="subsection.FS.SOL"></a> <a 
 id="x36-150000doc"></a><a 
 id="dx36-150001"></a> <a 
 id="solution.FS.C25"><span 
class="cmbx-12">C25</span></a>   Contributed&#x00A0;by&#x00A0;<a 
href="fcla-jsmath-latestli6.html#BeezerRobert">Robert&#x00A0;Beezer</a>    <a 
href="#exercise.FS.C25">Statement</a>&#x00A0;[<a 
href="#x36-149000doc">826<!--tex4ht:ref: exercise.FS.C25 --></a>]<br 
class="newline" />Add a <!--l. 10--><span class="math" 
>4 &#x00D7; 4</span> identity matrix
to the right of <!--l. 10--><span class="math" 
>A</span> to form
the matrix <!--l. 10--><span class="math" 
>M</span> and then
row-reduce to the matrix <!--l. 10--><span class="math" 
>N</span>,
</p><table class="equation-star"><tr><td>
<!--l. 12--><div class="math" 
>
M = \left [\array{ 
&#x2212;5&amp; 3 &amp;&#x2212;1&amp;1&amp;0&amp;0&amp;0\cr 
&#x2212;1 &amp; 1 &amp; 1 &amp;0 &amp;1 &amp;0 &amp;0
\cr 
&#x2212;8&amp; 5 &amp;&#x2212;1&amp;0&amp;0&amp;1&amp;0\cr 
 3 &amp;&#x2212;2 &amp; 0 &amp;0 &amp;0 &amp;0 &amp;1
   }                                                                                              \right ]\mathop{\longrightarrow}\limits_{}^{\text{RREF}}\left [\array{ 
\text{1}&amp;0&amp;2&amp;0&amp;0&amp;&#x2212;2&amp;&#x2212;5\cr 
0&amp;\text{1 } &amp;3 &amp;0 &amp;0 &amp;&#x2212;3 &amp;&#x2212;8
\cr 
0&amp;0&amp;0&amp;\text{1}&amp;0&amp;&#x2212;1&amp;&#x2212;1\cr 
0&amp;0 &amp;0 &amp;0 &amp;\text{1 } &amp; 1 &amp; 3 }                                                                                   \right ] = N
</div></td></tr></table>
<!--l. 32--><p class="indent" >   To apply <a 
href="#theorem.FS">Theorem&#x00A0;FS</a> in each of these four parts, we need the two
matrices,
</p><!--tex4ht:inline--><!--l. 46--><div class="math" 
>\eqalignno{
            C            &amp; = \left [\array{ 
\text{1}&amp;0&amp;2\cr 
0&amp;\text{1 } &amp;3}                                                                                              \right ]            &amp;L            &amp; = \left [\array{ 
\text{1}&amp;0&amp;&#x2212;1&amp;&#x2212;1\cr 
0&amp;\text{1 } &amp; 1 &amp; 3 }                                                                                        \right ]            &amp;            &amp;            &amp;            &amp;
   }</div>
<!--l. 48--><p class="noindent" >(a)
                                                                          

                                                                          
</p><!--tex4ht:inline--><!--l. 54--><div class="math" 
>\eqalignno{
               &#x211B;\kern -1.95872pt \left (A\right )               &amp; = &#x211B;\kern -1.95872pt \left (C\right )                                                                                                                                                              &amp;               &amp;\text{@(a 
href="#theorem.FS")Theorem FS@(/a)}                 &amp;               &amp;               &amp;               &amp;
               \cr 
                    &amp; = \left \langle \left [\array{ 
1\cr 
0
\cr 
2   }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
0\cr 
1
\cr 
3   }                                                                                              \right ]\right \rangle                &amp;               &amp;\text{@(a 
href="fcla-jsmath-latestli34.html#theorem.BRS")Theorem BRS@(/a)}               &amp;               &amp;               &amp;               &amp;
   }</div>
<!--l. 56--><p class="noindent" >(b)
</p><!--tex4ht:inline--><!--l. 62--><div class="math" 
>\eqalignno{
              C\kern -1.95872pt \left (A\right )              &amp; = N\kern -1.95872pt \left (L\right )                                                                                                                                                             &amp;              &amp;\text{@(a 
href="#theorem.FS")Theorem FS@(/a)}                &amp;              &amp;              &amp;              &amp;
              \cr 
                   &amp; = \left \langle \left [\array{ 
 1\cr 
&#x2212;1
\cr 
 1\cr 
 0  }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
 1\cr 
&#x2212;3
\cr 
 0\cr 
 1  }                                                                                              \right ]\right \rangle               &amp;              &amp;\text{@(a 
href="fcla-jsmath-latestli26.html#theorem.BNS")Theorem BNS@(/a)}              &amp;              &amp;              &amp;              &amp;
   }</div>
<!--l. 64--><p class="noindent" >(c)
                                                                          

                                                                          
</p><!--tex4ht:inline--><!--l. 70--><div class="math" 
>\eqalignno{
               N\kern -1.95872pt \left (A\right )               &amp; = N\kern -1.95872pt \left (C\right )                                                                                   &amp;               &amp;\text{@(a 
href="#theorem.FS")Theorem FS@(/a)}                 &amp;               &amp;               &amp;               &amp;
               \cr 
                     &amp; = \left \langle \left [\array{ 
&#x2212;2\cr 
&#x2212;3
\cr 
 1  }                                                                                              \right ]\right \rangle                &amp;               &amp;\text{@(a 
href="fcla-jsmath-latestli26.html#theorem.BNS")Theorem BNS@(/a)}               &amp;               &amp;               &amp;               &amp;
   }</div>
<!--l. 72--><p class="noindent" >(d)
</p><!--tex4ht:inline--><!--l. 78--><div class="math" 
>\eqalignno{
              &#x2112;\kern -1.95872pt \left (A\right )              &amp; = &#x211B;\kern -1.95872pt \left (L\right )                                                                                                                                                             &amp;              &amp;\text{@(a 
href="#theorem.FS")Theorem FS@(/a)}                &amp;              &amp;              &amp;              &amp;
              \cr 
                   &amp; = \left \langle \left [\array{ 
 1\cr 
 0
\cr 
&#x2212;1\cr 
&#x2212;1
   }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
0\cr 
1
\cr 
1\cr 
3   }                                                                                              \right ]\right \rangle               &amp;              &amp;\text{@(a 
href="fcla-jsmath-latestli34.html#theorem.BRS")Theorem BRS@(/a)}              &amp;              &amp;              &amp;              &amp;
   }</div>
<!--l. 11--><p class="noindent" ><a 
 id="solution.FS.C26"><span 
class="cmbx-12">C26</span></a>   Contributed&#x00A0;by&#x00A0;<a 
href="fcla-jsmath-latestli6.html#BeezerRobert">Robert&#x00A0;Beezer</a>    <a 
href="#exercise.FS.C26">Statement</a>&#x00A0;[<a 
href="#x36-149000doc">826<!--tex4ht:ref: exercise.FS.C26 --></a>]<br 
class="newline" />For both parts, we need the extended echelon form of the matrix.
                                                                          

                                                                          
</p><!--tex4ht:inline--><!--l. 27--><div class="math" 
>\eqalignno{
\left [\array{ 
&#x2212;7&amp;&#x2212;11&amp;&#x2212;19&amp;&#x2212;15&amp;1&amp;0&amp;0&amp;0\cr 
 6 &amp; 10 &amp; 18 &amp; 14 &amp;0 &amp;1 &amp;0 &amp;0
\cr 
 3 &amp;  5  &amp;  9  &amp;  7  &amp;0&amp;0&amp;1&amp;0\cr 
&#x2212;1 &amp; &#x2212;2 &amp; &#x2212;4 &amp; &#x2212;3 &amp;0 &amp;0 &amp;0 &amp;1 }                                                                        \right ]\mathop{\longrightarrow}\limits_{}^{\text{RREF}}\left [\array{ 
\text{1}&amp;0&amp;&#x2212;2&amp;&#x2212;1&amp;0&amp;0&amp; 2 &amp; 5\cr 
0&amp;\text{1 } &amp; 3 &amp; 2 &amp;0 &amp;0 &amp;&#x2212;1 &amp;&#x2212;3
\cr 
0&amp;0&amp; 0 &amp; 0 &amp;\text{1}&amp;0&amp; 3 &amp; 2\cr 
0&amp;0 &amp; 0 &amp; 0 &amp;0 &amp;\text{1 } &amp;&#x2212;2 &amp; 0 }                                                                              \right ]&amp;&amp;
   }</div>
<!--l. 29--><p class="noindent" >From this matrix we extract the last two rows, in the last four columns to form the
matrix <!--l. 29--><span class="math" 
>L</span>,
</p><!--tex4ht:inline--><!--l. 38--><div class="math" 
>\eqalignno{
                           L = \left [\array{ 
\text{1}&amp;0&amp; 3 &amp;2\cr 
0&amp;\text{1 } &amp;&#x2212;2 &amp;0 }                                                                                          \right ]                           &amp;                           &amp;
   }</div>
<!--l. 40--><p class="noindent" >(a)   By <a 
href="#theorem.FS">Theorem&#x00A0;FS</a> and <a 
href="fcla-jsmath-latestli26.html#theorem.BNS">Theorem&#x00A0;BNS</a> we have
</p><!--tex4ht:inline--><!--l. 47--><div class="math" 
>\eqalignno{
                       C\kern -1.95872pt \left (D\right ) = N\kern -1.95872pt \left (L\right ) = \left \langle \left \{\left [\array{ 
&#x2212;3\cr 
 2
\cr 
 1\cr 
 0  }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
&#x2212;2\cr 
 0
\cr 
 0\cr 
 1  }                                                                                              \right ]\right \}\right \rangle                        &amp;                       &amp;
   }</div>
                                                                          

                                                                          
<!--l. 49--><p class="noindent" >(b)   By <a 
href="#theorem.FS">Theorem&#x00A0;FS</a> and <a 
href="fcla-jsmath-latestli34.html#theorem.BRS">Theorem&#x00A0;BRS</a> we have
</p><!--tex4ht:inline--><!--l. 56--><div class="math" 
>\eqalignno{
                        &#x2112;\kern -1.95872pt \left (D\right ) = &#x211B;\kern -1.95872pt \left (L\right ) = \left \langle \left \{\left [\array{ 
1\cr 
0
\cr 
3\cr 
2   }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
 0\cr 
 1
\cr 
&#x2212;2\cr 
 0 }                                                                                              \right ]\right \}\right \rangle                         &amp;                        &amp;
   }</div>
<!--l. 12--><p class="noindent" ><a 
 id="solution.FS.C60"><span 
class="cmbx-12">C60</span></a>   Contributed&#x00A0;by&#x00A0;<a 
href="fcla-jsmath-latestli6.html#BeezerRobert">Robert&#x00A0;Beezer</a>    <a 
href="#exercise.FS.C60">Statement</a>&#x00A0;[<a 
href="#x36-149000doc">828<!--tex4ht:ref: exercise.FS.C60 --></a>]<br 
class="newline" />(a)   The definition of the column space is the span of the set of columns
(<a 
href="fcla-jsmath-latestli34.html#definition.CSM">Definition&#x00A0;CSM</a>). So the desired set is just the four columns of
<!--l. 10--><span class="math" 
>B</span>,
</p><table class="equation-star"><tr><td>
<!--l. 12--><div class="math" 
>
S = \left \{\left [\array{ 
 2\cr 
 1
\cr 
&#x2212;1 }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
3\cr 
1
\cr 
2   }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
1\cr 
0
\cr 
3   }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
 1\cr 
 1
\cr 
&#x2212;4 }                                                                                              \right ]\right \}
</div></td></tr></table>
<!--l. 22--><p class="indent" >   (b)   <a 
href="fcla-jsmath-latestli34.html#theorem.BCS">Theorem&#x00A0;BCS</a> suggests row-reducing the matrix and using the columns
of <!--l. 23--><span class="math" 
>B</span>
that correspond to the pivot columns. </p><table class="equation-star"><tr><td>
                                                                          

                                                                          
<!--l. 25--><div class="math" 
>
B\mathop{\longrightarrow}\limits_{}^{\text{RREF}}\left [\array{ 
\text{1}&amp;0&amp;&#x2212;1&amp; 2\cr 
0&amp;\text{1 } &amp; 1 &amp;&#x2212;1
\cr 
0&amp;0&amp; 0 &amp; 0 }                                                                                         \right ]
</div></td></tr></table>
<!--l. 34--><p class="indent" >   So the pivot columns are numbered by elements of
<!--l. 34--><span class="math" 
>D = \left \{1,\kern 1.95872pt 2\right \}</span>, so
the requested set is </p><table class="equation-star"><tr><td>
<!--l. 36--><div class="math" 
>
S = \left \{\left [\array{ 
 2\cr 
 1
\cr 
&#x2212;1 }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
3\cr 
1
\cr 
2   }                                                                                              \right ]\right \}
</div></td></tr></table>
<!--l. 42--><p class="indent" >   (c)   We can find this set by row-reducing the transpose of
<!--l. 43--><span class="math" 
>B</span>,
deleting the zero rows, and using the nonzero rows as column vectors in the set.
This is an application of <a 
href="fcla-jsmath-latestli34.html#theorem.CSRST">Theorem&#x00A0;CSRST</a> followed by <a 
href="fcla-jsmath-latestli34.html#theorem.BRS">Theorem&#x00A0;BRS</a>.
</p><table class="equation-star"><tr><td>
                                                                          

                                                                          
<!--l. 45--><div class="math" 
>{
B}^{t}\mathop{\longrightarrow}\limits_{}^{\text{RREF}}\left [\array{ 
\text{1}&amp;0&amp; 3\cr 
0&amp;\text{1 } &amp;&#x2212;7
\cr 
0&amp;0&amp; 0\cr 
0&amp;0 &amp; 0 }                                                                                             \right ]
</div></td></tr></table>
<!--l. 55--><p class="indent" >   So the requested set is </p><table class="equation-star"><tr><td>
<!--l. 57--><div class="math" 
>
S = \left \{\left [\array{ 
1\cr 
0
\cr 
3   }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
 0\cr 
 1
\cr 
&#x2212;7 }                                                                                              \right ]\right \}
</div></td></tr></table>
<!--l. 64--><p class="indent" >   (d)   With the column space expressed as a null space, the vectors obtained
via <a 
href="fcla-jsmath-latestli26.html#theorem.BNS">Theorem&#x00A0;BNS</a> will be of the desired shape. So we first proceed with
<a 
href="#theorem.FS">Theorem&#x00A0;FS</a> and create the extended echelon form, </p><table class="equation-star"><tr><td>
<!--l. 67--><div class="math" 
>
\left [\left .B\kern 1.95872pt \right \vert \kern 1.95872pt {I}_{3}\right ] = \left [\array{ 
 2 &amp;3&amp;1&amp; 1 &amp;1&amp;0&amp;0\cr 
 1 &amp;1 &amp;0 &amp; 1 &amp;0 &amp;1 &amp;0
\cr 
&#x2212;1&amp;2&amp;3&amp;&#x2212;4&amp;0&amp;0&amp;1 }                                                                                   \right ]\mathop{\longrightarrow}\limits_{}^{\text{RREF}}\left [\array{ 
\text{1}&amp;0&amp;&#x2212;1&amp; 2 &amp;0&amp; {2\over  
3}  &amp;{&#x2212;1\over 
 3}
\cr 
0&amp;\text{1}&amp; 1 &amp;&#x2212;1&amp;0&amp; {1\over  
3}  &amp;  {1\over  
3}
\cr 
0&amp;0&amp; 0 &amp; 0 &amp;\text{1}&amp;{&#x2212;7\over 
 3}  &amp;{&#x2212;1\over 
 3} }                                                                                                               \right ]
</div></td></tr></table>
<!--l. 81--><p class="indent" >   So, employing <a 
href="#theorem.FS">Theorem&#x00A0;FS</a>, we have
<!--l. 81--><span class="math" 
>C\kern -1.95872pt \left (B\right ) = N\kern -1.95872pt \left (L\right )</span>,
where </p><table class="equation-star"><tr><td>
                                                                          

                                                                          
<!--l. 83--><div class="math" 
>
L = \left [\array{ 
\text{1}&amp;{&#x2212;7\over 
 3}  &amp;{&#x2212;1\over 
 3} }                                                                                                                              \right ]
</div></td></tr></table>
<!--l. 90--><p class="indent" >   We can find the desired set of vectors from <a 
href="fcla-jsmath-latestli26.html#theorem.BNS">Theorem&#x00A0;BNS</a> as </p><table class="equation-star"><tr><td>
<!--l. 92--><div class="math" 
>
S = \left \{\left [\array{ 
{7\over 
3}
\cr 
1\cr 
0   }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
{1\over 
3}
\cr 
0\cr 
1   }                                                                                              \right ]\right \}
</div></td></tr></table>
<!--l. 13--><p class="noindent" ><a 
 id="solution.FS.C61"><span 
class="cmbx-12">C61</span></a>   Contributed&#x00A0;by&#x00A0;<a 
href="fcla-jsmath-latestli6.html#BeezerRobert">Robert&#x00A0;Beezer</a>    <a 
href="#exercise.FS.C61">Statement</a>&#x00A0;[<a 
href="#x36-149000doc">829<!--tex4ht:ref: exercise.FS.C61 --></a>]<br 
class="newline" />(a)   First find a matrix <!--l. 10--><span class="math" 
>B</span>
that is row-equivalent to <!--l. 10--><span class="math" 
>A</span>
and in reduced row-echelon form </p><table class="equation-star"><tr><td>
<!--l. 12--><div class="math" 
>
B = \left [\array{ 
\text{1}&amp;0&amp;3&amp;&#x2212;2\cr 
0&amp;\text{1 } &amp;1 &amp;&#x2212;1
\cr 
0&amp;0&amp;0&amp; 0 }                                                                                           \right ]
</div></td></tr></table>
                                                                          

                                                                          
<!--l. 21--><p class="indent" >   By <a 
href="fcla-jsmath-latestli34.html#theorem.BCS">Theorem&#x00A0;BCS</a> we can choose the columns of
<!--l. 21--><span class="math" 
>A</span> that correspond to
dependent variables (<!--l. 21--><span class="math" 
>D = \left \{1, 2\right \}</span>)
as the elements of <!--l. 21--><span class="math" 
>S</span>
and obtain the desired properties. So </p><table class="equation-star"><tr><td>
<!--l. 23--><div class="math" 
>
S = \left \{\left [\array{ 
 2\cr 
&#x2212;5
\cr 
 1  }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
&#x2212;1\cr 
 3
\cr 
 1  }                                                                                              \right ]\right \}
</div></td></tr></table>
<!--l. 27--><p class="indent" >   (b)   We can write the column space of
<!--l. 27--><span class="math" 
>A</span> as the
row space of the transpose (<a 
href="fcla-jsmath-latestli34.html#theorem.CSRST">Theorem&#x00A0;CSRST</a>). So we row-reduce the transpose of
<!--l. 27--><span class="math" 
>A</span> to obtain the
row-equivalent matrix <!--l. 27--><span class="math" 
>C</span>
in reduced row-echelon form </p><table class="equation-star"><tr><td>
<!--l. 29--><div class="math" 
>
C = \left [\array{ 
1&amp;0&amp;8\cr 
0&amp;1 &amp;3
\cr 
0&amp;0&amp;0\cr 
0&amp;0 &amp;0}                                                                                              \right ]
</div></td></tr></table>
<!--l. 39--><p class="indent" >   The nonzero rows (written as columns) will be a linearly independent set that spans the
row space of <!--l. 39--><span class="math" 
>{A}^{t}</span>,
by <a 
href="fcla-jsmath-latestli34.html#theorem.BRS">Theorem&#x00A0;BRS</a>, and the zeros and ones will be at the top of the vectors,
                                                                          

                                                                          
</p><table class="equation-star"><tr><td>
<!--l. 41--><div class="math" 
>
S = \left \{\left [\array{ 
1\cr 
0
\cr 
8   }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
0\cr 
1
\cr 
3   }                                                                                              \right ]\right \}
</div></td></tr></table>
<!--l. 45--><p class="indent" >   (c)   In preparation for <a 
href="#theorem.FS">Theorem&#x00A0;FS</a>, augment
<!--l. 45--><span class="math" 
>A</span> with the
<!--l. 45--><span class="math" 
>3 &#x00D7; 3</span> identity
matrix <!--l. 45--><span class="math" 
>{I}_{3}</span>
and row-reduce to obtain the extended echelon form, </p><table class="equation-star"><tr><td>
<!--l. 47--><div class="math" 
>
\left [\array{ 
1&amp;0&amp;3&amp;&#x2212;2&amp;0&amp;&#x2212;{1\over 
8}&amp;  {3\over  
8}
\cr 
0&amp;1&amp;1&amp;&#x2212;1&amp;0&amp; {1\over  
8}  &amp;  {5\over  
8}
\cr 
0&amp;0&amp;0&amp; 0 &amp;1&amp; {3\over  
8}  &amp;&#x2212;{1\over 
8} }                                                                                                               \right ]
</div></td></tr></table>
<!--l. 55--><p class="indent" >   Then since the first four columns of row 3 are all zeros, we extract
</p><table class="equation-star"><tr><td>
                                                                          

                                                                          
<!--l. 57--><div class="math" 
>
L = \left [\array{ 
\text{1}&amp;{3\over 
8}&amp;&#x2212;{1\over 
8} }                                                                                                                              \right ]
</div></td></tr></table>
<!--l. 64--><p class="indent" >   <a 
href="#theorem.FS">Theorem&#x00A0;FS</a> says that <!--l. 64--><span class="math" 
>C\kern -1.95872pt \left (A\right ) = N\kern -1.95872pt \left (L\right )</span>.
We can then use <a 
href="fcla-jsmath-latestli26.html#theorem.BNS">Theorem&#x00A0;BNS</a> to construct the desired set
<!--l. 64--><span class="math" 
>S</span>, based on the free variables
with indices in <!--l. 64--><span class="math" 
>F = \left \{2, 3\right \}</span> for the
homogeneous system <!--l. 64--><span class="math" 
>&#x2112;S\kern -1.95872pt \left (L,\kern 1.95872pt 0\right )</span>,
so </p><table class="equation-star"><tr><td>
<!--l. 66--><div class="math" 
>
S = \left \{\left [\array{ 
&#x2212;{3\over 
8}
\cr 
1\cr 
 0  }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
{1\over 
8}
\cr 
0\cr 
1   }                                                                                              \right ]\right \}
</div></td></tr></table>
<!--l. 70--><p class="indent" >   Notice that the zeros and ones are at the bottom of the vectors.<br 
class="newline" />(d)   This is a straightforward application of <a 
href="fcla-jsmath-latestli34.html#theorem.BRS">Theorem&#x00A0;BRS</a>. Use the row-reduced
matrix <!--l. 72--><span class="math" 
>B</span>
from part (a), grab the nonzero rows, and write them as column vectors,
</p><table class="equation-star"><tr><td>
                                                                          

                                                                          
<!--l. 74--><div class="math" 
>
S = \left \{\left [\array{ 
 1\cr 
 0
\cr 
 3\cr 
&#x2212;2 }                                                                                              \right ],\kern 1.95872pt \left [\array{ 
 0\cr 
 1
\cr 
 1\cr 
&#x2212;1 }                                                                                              \right ]\right \}
</div></td></tr></table>
                                                                          

                                                                          
   <h4 class="likesubsectionHead"><a 
 id="x36-151000"></a>Annotated Acronyms M: Matrices</h4>
<!--l. 394--><p class="noindent" ><a 
 id="acronyms.M"></a> <a 
 id="x36-151000doc"></a><a 
 id="dx36-151001"></a> <a 
href="fcla-jsmath-latestli30.html#theorem.VSPM">Theorem&#x00A0;VSPM</a><br 
class="newline" />These are the fundamental rules for working with the addition, and scalar
multiplication, of matrices. We saw something very similar in the previous chapter
(<a 
href="fcla-jsmath-latestli23.html#theorem.VSPCV">Theorem&#x00A0;VSPCV</a>). Together, these two definitions will provide our definition for
the key definition, <a 
href="fcla-jsmath-latestli37.html#definition.VS">Definition&#x00A0;VS</a>.
</p><!--l. 23--><p class="noindent" ><a 
href="fcla-jsmath-latestli31.html#theorem.SLEMM">Theorem&#x00A0;SLEMM</a><br 
class="newline" /><a 
href="fcla-jsmath-latestli24.html#theorem.SLSLC">Theorem&#x00A0;SLSLC</a> connected linear combinations with systems of equations.
<a 
href="fcla-jsmath-latestli31.html#theorem.SLEMM">Theorem&#x00A0;SLEMM</a> connects the matrix-vector product (<a 
href="fcla-jsmath-latestli31.html#definition.MVP">Definition&#x00A0;MVP</a>) and
column vector equality (<a 
href="fcla-jsmath-latestli23.html#definition.CVE">Definition&#x00A0;CVE</a>) with systems of equations. We&#x2019;ll see this
one regularly.
</p><!--l. 27--><p class="noindent" ><a 
href="fcla-jsmath-latestli31.html#theorem.EMP">Theorem&#x00A0;EMP</a><br 
class="newline" />This theorem is a workhorse in <a 
href="fcla-jsmath-latestli31.html#section.MM">Section&#x00A0;MM</a> and will continue to make regular
appearances. If you want to get better at formulating proofs, the application of
this theorem can be a key step in gaining that broader understanding. While it
might be hard to imagine <a 
href="fcla-jsmath-latestli31.html#theorem.EMP">Theorem&#x00A0;EMP</a> as a <span 
class="cmti-12">definition </span>of matrix multiplication,
we&#x2019;ll see in <a 
href="fcla-jsmath-latestli57.html#exercise.MR.T80">Exercise&#x00A0;MR.T80</a> that in theory it is actually a <span 
class="cmti-12">better </span>definition of
matrix multiplication long-term.
</p><!--l. 31--><p class="noindent" ><a 
href="fcla-jsmath-latestli32.html#theorem.CINM">Theorem&#x00A0;CINM</a><br 
class="newline" />The inverse of a matrix is key. Here&#x2019;s how you can get one if you know how to
row-reduce.
</p><!--l. 35--><p class="noindent" ><a 
href="fcla-jsmath-latestli33.html#theorem.NI">Theorem&#x00A0;NI</a><br 
class="newline" />&#x201C;Nonsingularity&#x201D; or &#x201C;invertibility&#x201D;? Pick your favorite, or show your versatility by
using one or the other in the right context. They mean the same thing.
</p><!--l. 39--><p class="noindent" ><a 
href="fcla-jsmath-latestli34.html#theorem.CSCS">Theorem&#x00A0;CSCS</a><br 
class="newline" />Given a coefficient matrix, which vectors of constants create consistent systems.
This theorem tells us that the answer is exactly those column vectors in the
column space. Conversely, we also use this teorem to test for membership in the
column space by checking the consistency of the appropriate system of
equations.
                                                                          

                                                                          
</p><!--l. 43--><p class="noindent" ><a 
href="fcla-jsmath-latestli34.html#theorem.BCS">Theorem&#x00A0;BCS</a><br 
class="newline" />Another theorem that provides a linearly independent set of vectors whose span
equals some set of interest (a column space this time).
</p><!--l. 47--><p class="noindent" ><a 
href="fcla-jsmath-latestli34.html#theorem.BRS">Theorem&#x00A0;BRS</a><br 
class="newline" />Yet another theorem that provides a linearly independent set of vectors whose
span equals some set of interest (a row space).
</p><!--l. 51--><p class="noindent" ><a 
href="fcla-jsmath-latestli34.html#theorem.CSRST">Theorem&#x00A0;CSRST</a><br 
class="newline" />Column spaces, row spaces, transposes, rows, columns. Many of the connections
between these objects are based on the simple observation captured in this
theorem. This is not a deep result. We state it as a theorem for convenience, so we
can refer to it as needed.
</p><!--l. 55--><p class="noindent" ><a 
href="#theorem.FS">Theorem&#x00A0;FS</a><br 
class="newline" />This theorem is inherently interesting, if not computationally satisfying. Null
space, row space, column space, left null space &#x2014; here they all are, simply by row
reducing the extended matrix and applying <a 
href="fcla-jsmath-latestli26.html#theorem.BNS">Theorem&#x00A0;BNS</a> and <a 
href="fcla-jsmath-latestli34.html#theorem.BCS">Theorem&#x00A0;BCS</a>
twice (each). Nice.
                                                                          

                                                                          
</p>
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